Post was placed on crawlspace vent cutout

Question: I'm a first time home buyer doing a new kitchen addition. The structural engineer drew out a framing plan where he put posts over a crawl space vent cutout. See photo for full detail. This is giving me a lot of anxiety and my GC for some reason has not flagged this either. This problem was not caught in plan check either. I'm  submitting amendments, so not sure if this was the case in the original plan or if my GC has built it according to original for inspection purposes and will fix it once the amendment is approved. Any advice on what I should do?

This:

and this:
Comment: The leftover joists plus the bottom plate plate remind me a lot of a double top plate, where double top plates accept roof rafter weights just fine. These "posts" are too nondescript. What loads do they carry? Is there a hip roof or something? Because I see joists running parallel with the notched joists, where that makes me think "nonbearing wall."
Comment: Are vents shown in the engineer's drawings? If so, you can be assured that the structure adequately supports the loads above, unless the engineer likes losing their license. If not, who put them there?
Comment: You need to communicate with your GC and then possibly the engineer.
Comment: Added another photo of the full vertical length of the post. The top is being built as a deck and needs to support active loads (including people). From what I can tell from the original plan and the amendment plan is that the vent spaces were moved in he drawings but they were already cut out by the GC as he was going by the original plan and not the amendment. Can he close and move the vent now or will a lot of things need to be torn down?
Answer: Worst case scenario, I imagine installing blocking to restore the load path. Rather than making something to fill the existing gap, you would probably cut away the remaining joist material above the notches and install new chunks of 2x10 or whatever that is.
Those joists above the foundation are technically part of the lateral system, but an interruption at such blocking isn't a big deal. I guess the floor sheathing fasteners won't get typical embedment in the underlying blocking with the bottom plate in the way, so the fastener schedule may be unorthodox.


Integration of System Tests in the build process

Question: I am continuing the development of a serialization layer generator. The user enters a description of types (currently in XSD or in WSDL), and the software produces code in a certain target language (currently, Java and ansi C89) which is able to represent the types described and which is also able to serialize (turn into a byte-sequence) and deserialize these values.
Since generating code is tricky (I mean, writing code is hard. Writing code that writes code is writing code to do a hard thing, which is a whole new land of hardness :) ). Thus, in the project which preceded my master thesis, we decided that we want some system tests in place.
These system tests know a type and a number of pairs of values and byte sequences. In order to execute a system test in a certain language, the type is run through the syste, resulting in code as described above. This code is then linked with some handwritten host-code, which is capable of reading these pairs of a byte sequence and a value and functions to read values of the given value from a string. The resulting executable is then run and the byte-value-pairs are fed into this executable and it is overall checked if all such bindings result in the output "Y". If this is the case, then these example values for the type serialize into the previously defined byte sequence and we can conclude that the generated code compiles and runs correctly, and thus, overall, that the part of the system handling this type is correct.  This is a very good thing.
However, right now I am a bit unhappy with the current implementation. Currently, I have written a custom junit runner which uses quite a lot of reflection sorcery in order to read these byte-value-bindings from a classes attributes. Also, the overall stack to generate the code requires a lot of boilerplate code and boilerplate classes which do little more than to contain two or three strings. Even worse, it is quite hard to get a good integration with all tools which base on Junits descriptions and which generate test failure reports. It is quite hard to actually debug what is happening if the helpful maven Junit testrunner or the eclipse test runner gobble up whatever errors the compiler threw, just because the format of this error is different from junits own assertion errors.
Even worse, a single failed test in the generated code causes the maven build to fail. This is very annoying. I like it if the maven build fails if a certain test of a different unit fails, because (for example), if a certain depth first preorder calculation fails for some reason, everything will go haywire. However, if I just want to show someone some generated code for a type I know working, then it is very annoying if I cannot quickly build my application because the type I am working on right now is not finished.
So, given this background, how can I get a nice automated system which checks these generation specifications? Possibilities I have considererd:

A Junit integrated solution appears to be less than ideal, unless I can improve the integration of maven and junit and junit with my runner and everything else.
We used fitnesse earlier, but overall ditched it, because it caused more problems than it solved. The major issues we had were integration into maven and hudson.
A solution using texttest. I am not entirely convinced, because this mostly wants an executable, strings to put on stdin and strings to expect on stdout. Adding the whole "run application, link with host code and THEN run the generated executable" seems kinda complicated.
Writing my own solution. This will of course work and do what I want. However, this will be the most time consuming task, as usual.

So... do you see another possible way to do this while avoiding to write something own?
Answer: You can run Maven with -Dmaven.test.skip=true. Netbeans has a way to set this automatically unless you explicitly hit one of the commands to test the project, I don't know about Eclipse.


SSH key for HTTP

Question: Is it possible to set something up like SSH Keys for http traffic, so that a browser must have the matching key in order to access?
Comment: Obviously it would have to be HTTPS, not HTTP, since HTTP doesn't involve any cryptography.
Answer: What you are looking for is called Mutual SSL Authentication.
In the case of a commonly deployed web server as Apache, and an also commonly deployed SSL/TLS implementation as OpenSSL, the steps would be as follows:
There is a handy script distributed alongside openssl, CA.sh to do most of this stuff. Its location is distribution specific. In Debian and derivatives you can locate it using:
# apt-file search CA.sh
openssl: /usr/lib/ssl/misc/CA.sh

And RedHat and derivatives the (approximate) equivalent is:
# yum provides */CA
1:openssl-1.0.1e-4.fc18.x86_64 : Utilities from the general purpose cryptography library with TLS implementation
Repo        : @updates
Matched from:
Filename    : /etc/pki/tls/misc/CA

This is a very simple bash script that eases the creation of the directory structure needed to manage a CA (this is described in the [ CA_default ] section of openssl.cnf). I recommend you to use it and look at the code to learn what it is actually doing.
# ./CA.sh -help
usage: ./CA.sh -newcert|-newreq|-newreq-nodes|-newca|-sign|-verify

This script will use the defaults provided in openssl.cnf, and/or you can provide one configuration file as an argument to the openssl $command using the -config switch if not using CA.sh.
The location of the openssl.cnf file is also distribution specific, and you can use the same commands above to find it. The one you want is the one provided by the openssl package.
You probably want to modify the following sections:
[ CA_default ]
default_days    = 365                   # how long to certify for
default_crl_days= 30                    # how long before next CRL

[ req ]
default_bits            = 2048

[ req_distinguished_name ]
countryName                     = Country Name (2 letter code)
countryName_default             = AU
countryName_min                 = 2
countryName_max                 = 2
stateOrProvinceName             = State or Province Name (full name)
stateOrProvinceName_default     = Some-State
localityName                    = Locality Name (eg, city)
0.organizationName              = Organization Name (eg, company)
0.organizationName_default      = Internet Widgits Pty Ltd
#1.organizationName             = Second Organization Name (eg, company)
#1.organizationName_default     = World Wide Web Pty Ltd
organizationalUnitName          = Organizational Unit Name (eg, section)
#organizationalUnitName_default =
commonName                      = Common Name (e.g. server FQDN or YOUR name)
commonName_max                  = 64
emailAddress                    = Email Address
emailAddress_max                = 64

Generate the CA

Once you have edited openssl.cnf to match your needs, you can generate a CA certificate. Depending on whether you need this CA signed by an recognized third-party CA or not, you can generate a self-signed CA or a CSR to be submitted for signing.
./CA.sh -newca
You will be interactively prompted to ask some questions, the defaults will appear between square brackets. You will recognize the options you modified in openssl.cnf here.

Generate a certificate for the server

You can create a certificate request for the server using the same script:
./CA.sh -newreq
Again, you will be prompted to ask several questions, the most important is the Common Name of the certificate, which must match the DNS resolvable name for the IP of the server (or you can use other means, like /etc/hosts, not recommended, hard to maintain and scale)
What you will obtain is a Certificate Signing Request (CSR for short). This will be signed by the Certificate Authority (CA) you created before.
./CA.sh -sign

Generate certificates for the clients

Repeat the steps of creating a CSR and getting it signed by the CA. When doing so, pay close attention to how you fill in the Common Name, Organization and Organizational Unit fields, as those will be required afterwards in order to configure the server.
A neat way to distribute the client certificates alongside their respective private keys and the CA certificate is using p12 bundles:
openssl pkcs12 -export -in Certificates/client.pem -inkey client.key -certfile CA.pem
-out clientcert.p12

Configure the server

Let's suppose the server you are referring to is an Apache web server. Once you have the server certificate, you configure the appropriate VHOST to serve whatever content is going to be protected by mutual SSL authentication. An example could be this phpmyadmin virtual hosts. This is working in an Apache 2.4 server, so please don't use it as-is and carefully review and test it to adapt it to your needs.
Listen                   443 https

<VirtualHost 184.108.40.206:443>
  DocumentRoot           "/srv/www/html"
  ServerAdmin            firstname.lastname@example.com
  SSLCACertificateFile   /etc/pki/CA/cacert.pem
  SSLCertificateFile     /etc/pki/tls/private/company.com/newcert.pem
  SSLCertificateKeyFile  /etc/pki/tls/private/company.com/newkey.pem
  SSLCARevocationCheck   chain
  SSLCARevocationFile    /etc/pki/CA/crl/crl.pem
  SSLEngine              on
  SSLStrictSNIVHostCheck on
  SSLVerifyClient        require
  SSLVerifyDepth         5
  ServerName             phpmyadmin.company.com
  RewriteEngine          on
  RewriteCond            %{REMOTE_ADDR} !^127\.0\.0\.1$
  RewriteCond            %{HTTPS} !=on
  RewriteRule            . - [F]
  ErrorLog               "|/usr/sbin/rotatelogs -L /var/log/httpd/foo/error.log -f /var/log/httpd/foo/error.log.%Y%m%d 86400"
  CustomLog              "|/usr/sbin/rotatelogs -L /var/log/httpd/foo/access.log -f /var/log/httpd/foo/access.log.%Y%m%d 86400" logstash_json
  <Directory /usr/share/bar/>
    Require              ssl
    Require              ssl-verify-client
    SSLRequireSSL
    SSLOptions           +FakeBasicAuth +StrictRequire
    SSLRequire           %{SSL_CIPHER_USEKEYSIZE} >= 256
    SSLRequire           %{SSL_CLIENT_S_DN_O} eq "Awesome Company" \
                     and %{SSL_CLIENT_S_DN_OU} eq "Development" \
                     and %{SSL_CLIENT_S_DN_CN} in {"John Doe", "Jane Doe"}
    SSLRenegBufferSize   131072
  </Directory>
</VirtualHost>

You can use as many per-directory access control as you need, the important part is that the client certificates shown must comply with the restrictions imposed by the SSLRequire directives, i.e., they must match the Organization, Organizational Unit and Common Name conditions (or other fields of the certificate as you see fit). These fields are taken from the client certificates.

Generate a Certificate Revocation List

In order to be able to revoke access to a client certificate, you need to generate a CRL.
The command to do it (provided you in the top of the CA directory structure):
openssl ca -config /path/to/openssl.cnf -gencrl -out crl/crl.pem

Afterwards, you revoke client certs as needed using:
openssl ca -config /path/to/openssl.cnf -revoke clientcert.pem

References:
OpenSSL online documentation and Apache online documentation.
Answer: Yes, it is called "Client Certificate Authentication" and is oft used with Smart Cards and the like.
It can be complicated to get running well, depending on what your desired product is.

SSH key for HTTP


relationship between eigenvalues/eigenvectors of A, B and AB

Question: What is the eigenvalue/eigenvector relationship between matrix A,B and AB?
Answer: I do not know about eigenvectors, but for the eigenvalues this is a special case of Deligne-Simpson Problem. It was completely solved by Crawley-Boevey about 10 years ago using quivers. For details, see my answer here and references therein.
Answer: I assume this is over $\mathbb C$.
If they don't commute, about all you can say is that the determinant (which is the product of the eigenvalues, counted by algebraic multiplicity) of $AB$ is the product of the determinants of $A$ and $B$.
Answer: See the following paper. The review describes only results for $A+B$, but this readily transforms to $AB$, as described in the paper.

MR1957068 (2004b:14093) Reviewed
Klyachko, Alexander(TR-BILK)
Vector bundles, linear representations, and spectral problems. (English summary) Proceedings of the International Congress of Mathematicians, Vol. II (Beijing, 2002), 599–613, Higher Ed. Press, Beijing, 2002.

Review: The text provides a review of the amazing development resulting from the ground-breaking discovery of the author [Selecta Math. (N.S.) 4 (1998), no. 3, 419–445; MR1654578 (2000b:14054)] of the fundamental connections between inequalities for eigenvalues of Hermitian operators $A$, $B$ and $A+B$, Littlewood-Richardson coefficients $c^\gamma_{\alpha\beta}$
for the decomposition of the tensor product of irreducible $GL_n$-representations
$$V_\alpha⊗V_β=∑c^γ_{αβ}V_γ,$$
the decomposition of the product of two Schubert cycles in the cohomology of the Grassmannian and toric stable vector bundles on P2. The topic has further ramifications for the groups $SL_n$, $SO_n$, $Sp_{2n}$ and other symmetric spaces. The paper contains no proofs but has an extensive list of references.
Comment: Peter: Klyachko indeed solved the additive problem, but only for symmetric/hermitian matrices. This translates to solution of multiplicative problem, but not for eigenvalues --- you have to use singular values instead. Belkale solved multiplicative problem for unitary matrices. Solution for any other real forms of $SL(n,C)$ is currently unknown.


minimum load current vs stability

Question: One of my previous classmates said "Low Drop Out regulator has a minimum load requirement and this is related to stability. It might affect start up overshoot." In the previous school project, my LDO, designed to produce a 1.8V regulated voltage, has a 0.6 V start up overshoot and that was the advice he gave me to improve my schematics by adding a resistor in parallel to my load to achieve some minimum load current even in zero load condition. It didn't improve anything though. He didn't explain further as he was in a rush then.
Today I recalled that statement and tried to search such statements on Internet and find a couple similar statements without explanation. I couldn't find that freindly classmate again as he graduated. Could someone give a detail explanation to me how does the relations of minimum load current and stability hold true?
Answer: In the design you show, the gm of the output stage will be dependent upon the load current and so will affect the loop gain and potentially stability. Providing a minimum load will reduce the variation in gm and allow a better compromise in the feedback compensation.
Why do you have a capacitor to ground at the output of the amplifier - that will tend to make stability worse by introducing a pole in the transfer function.
Many voltage regulators have minimum load constraints for a different reason. Three pin regulators such as LM317 intentionally divert most of their internal operating current to the output in order to keep the current into the adjust pin constant. This means that if the output current falls below the operating current of the internal circuitry LDO will lose control and the output voltage will rise.
Comment: Don't know how you ended up with "willies" in the last sentence there!! Is it supposed to read will lose?
Comment: *&^% autocorrect...

minimum load current vs stability


How to negotiate for higher salary

Question: I got a call from recruiter today to offer me 44,000 for a Financial analyst position. I asked for $48,000 and above in the interview. Told her to up it. She called back that the hiring manager said she cant go above 44,000. I would like to work at the company and wish they can truly increase the salary. How can i figure out whether  to accept or  try to re - negotiate or wait and hope they change their mind. They told me at the interview the highest range is 60,000
Comment: Either take the 44 or tell them you will walk away if they don't give you $X.
Comment: "How to negotiate" is a large enough topic that many books have been written about it. You are not going to be told the secrets in a three-paragraph answer here. Also neither you nor we know whether they might be prepared to pay $48,000 for you. There is no "These are not the droids you are looking for" trick you can do to change their mind.
Comment: Hey Bernice. Currently, this question is off-topic, as it is asking us to make a choice for you... please read [how to ask](https://workplace.stackexchange.com/help/dont-ask) and try to rephrase your post so it can be ontopic.
Comment: You already have your answer. You asked for what you want in a salary, they told you they can't give it to you. Only you can decide if you would rather work for your desired salary, or work for this company. No one here can help you with this.
Answer: Only you can determine what your acceptable salary is.  Do you believe you'd be happy making that much at that company?   Or do you believe you'd be better off waiting and hoping to get more at a different company?  Benefits also play into it.   Get information on the health insurance offered, if any, along with vacation time.   If you get 2 weeks of vacation and good insurance, that's easily worth $48 k without benefits in a different company.

How to negotiate for higher salary


Is it OK to promote linguistics.stackexchange.com here?

Question: Linguistics.se is nearing 100 % commitment (currently only 18 more people needed). Is it acceptable to bring this to the attention of people here, or is that considered spamming? I was hoping speakers of German might be interested, since the sites will be closely related:
Linguistics
Comment: It's nice of you to ask while posting the link :)
Comment: Heh. Yeah. *May I promote  on this site or is that considered spamming?* ;)
Comment: @TimN: I plead guilty! Though I was also genuinely interested in whether or not this spamming was appreciated, and that would have been hard without mentioning which specific proposal I had in mind...
Answer: imho it should always be OK if its likely that some questions get migrated from proposal to proposal. We already had questions here which got closed or would better fit linguistics.
I already asked here to commit to other proposals related to language.
I would recommend to place a link on all proposals/Betas related to language in each others sidebar on the main page, so we build a smaller language-network on SE and profitate a bit from the highly frequented english.SE. People dont look on and on what new Betas there are, or mainly frequent their top 3 sites and forget. I dont look very often on the main site stackexchange.com, too overloaded to me...
Comment: OK thanks, I agree. And putting the links in a sidebar would be an excellent idea!


Enchanting non-enchantable items on MCPE 1.2.3 iOS

Question: I know that there is the /enchant command, but it, as usual, only does armor and weapons and fishing rods, and my long-waited "Sticky the Stick" is unmakeable. Any suggestions on how to enchant any item without a special app? Commands only, please.
Answer: It looks like "non-enchantable" items cannot normally be crafted.
These bug reports denounce it.
Bug MCPE 15500
Bug MCPE 11993
However, using commands or creative mode, it is possible to do. Check out this link to another Arqade question.
From what I have seen, it is possible to use an anvil and enchanted book in creative mode. You will also be able to name your stick "Sticky the Stick".
You could also use the /give command to get an enchanted stick. Check out the Arqade question link earlier in this post for more information about this topic.
Note that /enchant will not work. This command only enchants "enchantable" items.
Update: I did not see the part about commands only. Sorry about that. I would recommend using an anvil and enchanted book in creative mode though since I know you can name it "Sticky the Stick".
Answer: No, there isn't, since the command makes sure to check if the enchantment is compatible with said item. So, if you have a sword, you couldn't give it protection, an armour enchantment. So, in turn, there aren't any enchantments made for basic items like food or sticks, leaving them without the ability to be enchanted.
Comment: Adding on to The Mattbat999, the `/enchant` command is restricted by what you can enchant in survival (or MCPE) on an anvil. On the Java Edition (PC or computer), you can enchant anything with an anvil if you're on creative or use the `/give` command. Related: https://gaming.stackexchange.com/questions/57717/how-do-i-enchant-un-enchantable-items


What are some authoritative/respected "best known implementation" websites/resources?

Question: EDIT:
Wow, the initial response to this question was quite negative. I think I might have triggered some pretty strong emotions by using the word "best"; it seems like a few people latched onto that word and decided to dismiss my question right away.
Obviously, there are many, many situations in which no single approach is "best", or at least, what ends up being the best solution to one problem will often not be the best solution for other, even similar, problems. I get that. But now let me try to elaborate on the reasoning behind what I'm actually asking.
I tend to find it easiest to explain myself using analogies, so here goes. In my current job I work almost exclusively in .NET. .NET has a lot of functionality built into the framework. A prime example is the System.Collections.Generic namespace, which has a bunch of collection classes that (almost) no .NET developer in his/her right mind would bother re-developing from scratch, because very good implementations are already there. If I am working on a problem that requires a doubly linked list, I'm not going to decide, "Okay, time to write a doubly linked list class"; I'm just going to use the LinkedList<T> that's already there, or, at most, extend it or wrap it with my own class that adds some extra functionality.
Am I saying the "best" version of a doubly linked list is LinkedList<T> from .NET? Of course not. That would be absurd. But I highly doubt .NET's implementation of LinkedList<T> is drastically different from most other established libraries' implementations of collections that are intended to serve the same purpose (that of a doubly linked list). On the other hand, I am relatively confident that if I were to write my own implementation from scratch, there'd be a considerable number of issues with it, in terms of robustness, performance, flexibility, etc. for one simple reason: not that I'm stupid, or lazy, or don't care about good code--simply that I'm one person, and I'm not an expert on linked lists, and I haven't thought of everything that needs to be taken into consideration when designing one.
But I happen to be a developer who does take an interest in how things are implemented internally. And so it would be nice if I could check out a page where some variant of a well thought-out design for a linked list--or for any fairly established concept for which robust, efficient implementations have been written--were available to view. (By the way, yes I am aware that the source code for .NET's LinkedList<T> is available. I'm just using that as an example; really I am talking about all problems with solutions for which good, working implementations exist.)
Now, I talked about this being something that is open; let me elaborate on that. I am not talking about sites like SourceForge.net, or CodePlex, or Google Code. These are all sites for hosting projects, i.e., applications or libraries tailored for some specific industry or field or otherwise categorizable purpose. What I'm talking about is something like this:
http://en.wikibooks.org/wiki/Category:Algorithms_and_data_structures
Maybe I should have just provided that link in the first place, as it probably illustrates what I'm getting at better than anything I've written so far. But I think the main point that differentiates what I'm asking about from any other site I've seen is that I was specifically wondering if there could be some way to work on a new problem--so, something for which there aren't necessarily any well-known, established implementations, again as in my linked list example--collaboratively, in a wiki-esque fashion, but not tied to any specific open-source project.
So, as a conclusion of sorts, I was kind of envisioning a situation like the following: I find myself faced with a new problem. Maybe it isn't common enough to be something that is addressed in a framework like .NET. But it's common enough that some developers here and there are independently working on it. If a website exists like what I'm imagining, maybe at some point one of those developers working on the problem could post an idea on that website, and over time others might discover it and suggest improvements/modifications, and given enough time and participation, a pretty darn good implementation might result from all this collaboration. And from there, eventually, something like this implementation might be considered fairly "standard", just like a linked list implementation, or a quicksort implementation, or, I don't know, some well-known pseudo-random number generator.
Does this make any more sense to anyone now? I feel quite confident that what I'm talking about is not absurd, but hey, if that's what people think, then maybe it is.

Open source projects are very popular. Some of these are libraries suited for specific purposes, the best of which include some very well-written code.
However, if you're interested in contributing to an open source project, finding a project that is well-suited to your skills can be quite a task. At the same time, if you're interesting in using an open source project in your own work, finding a project that is well-suited to your needs can also be difficult, especially when, for example, open-source library X has a lot of functionality you could use, as does library Y, and these two libraries' capabilities overlap so that integrating both into your code could be messy.
We've all seen questions, here on Stack Overflow and elsewhere on the web, posted by one developer: "How would I implement this idea?" and answered by others, often accompanied by a plethora of example code. Sometimes these answers link to an open source project/library that provides functionality similar to what the poster is asking about.
My question is: are there any well-known websites or other sources that are open in nature and provide "best-known implementations" for common (or even not-so-common) programming problems, but not associated with any particular open source project?
As a generic example, suppose I have a need for some algorithm that does X. I post a question on SO or some other site requesting ideas, asking for suggestions on how best to implement it. One person points me to project P1, which contains some code that performs something very similar to this algorithm. Another person points me to project P2. Someone else writes some sample code and says, "maybe you could do it like this."
It seems to me, if there are all these different versions of this idea floating around out in the world, it would make sense for there to be a site, somewhat in the vein of Wikipedia, where a quasi-"official" implementation ("official" is not the right word; I'm just having trouble thinking of a better one right now) could be published and modified as improvements are developed/discovered.
I feel like I have stumbled across a few different sites like this in the past, but I'm interested to know if anyone else has found any resources like what I'm describing.
Comment: Wouldn't it be great if there was one authoritative, final repository for all good ideas.  Keep hoping.  And while your at it, remind people to stop improving on other people's ideas because new things keep fragmenting the One Repository for Good Ideas.
Comment: SourceForge is not a repository for "best of" projects.  It is a repository for **any** project.
Comment: @S.Lott: He did say **best known.**
Comment: IME, most of the best ideas, algorithms, code samples and open source projects tend to bubble up to the top of a Google search anyway.
Comment: Isn't this exactly what StackOverflow aspires to be?  If you want to know about the best implementation of an algorithm, just post a question and see what gets voted up... and then make your own decision.
Comment: @Robert: That may be, but the *main* point I was interested in was finding a site that involves collaboration and allows people to make updates to it, much like Wikipedia. Of course it's nice to know how established, existing algorithms/structures are commonly implemented, but I'm just as interested in solutions to new problems that are actually in the *process* of being developed.
Comment: @Daniel: Is it? I was kind of under the impression Stack Overflow is more about answering **any** programming-related question, and not necessarily providing good working implementations of solutions to new problems. It also seems that often when a question *is* posed as "How would I implement this?" the best answer ends up being a rough sketch, or a conceptual outline of what the solution might be, not a concrete implementation (though these are occasionally offered as well).
Comment: -1: After the edit, I no longer get the question.  It's sort of a rant, or a wish for collaborative development of algorithms.  But it's not a question that I can parse.
Comment: Too broad a question. SO is for well defined programming questions. How do I do x with tool y? Are you looking for good learning resources? Ask about that!
Answer: The very idea is absurd.  It means that there's one, single opinion on "best-known implementations" with no changes based on other people having better ideas.
It implies a that best practices are static and can be accumulated into a single repository.
If they could be collected, then Google would have them and would simply charge for access.
Interestingly, they don't have all the best practices.  Interestingly, they have to expend mountains of computing power looking for more information.  Then people (like you) have to read and think and judge and decide.
The read-think-judge-decide is really hard to eliminate.  Unless, of course, you want someone to think for you.  In which case, there are many companies who have a single solution that requires less thinking.  Call Microsoft or Oracle or IBM.  They have solutions that are all in one place, unified best practices, no reading, no thinking, no judging, no deciding required.
Open -- by definition -- means it's impossible to have a single authoritative source.
Comment: True, but you probably could come up with something like "a pretty good implementation that a lot of people agree is suitable for use case".    Sure, it's not BEST, but still useful.
Comment: Google wouldn't charge for access.  They'd just display ads.
Comment: @S.Lott: really, *absurd*? That sounds like a knee-jerk reaction to me. To be fair, I did use some poorly chosen words in my original question, namely "best", which I didn't really mean in the objective sense in which you interpreted it. Still, I think you're kind of unfairly representing my question in this response. Anyway, I expanded on my original post, and would be interested to hear any further thoughts you may have (even though originally I was just asking for links, in case people knew of any sites like what I was describing).
Comment: @Dan: Not knee-jerk.  Just logic.  "Single Best" (or "Standard") and "Innovation" are diametrically opposed.  Period.  The "single best" networking protocol for information sharing can't be HTTP -- it's awful.  But there it is.  I can't figure out what you're asking for.
Comment: @S.Lott: The original question was pretty straightforward: does anyone know of any sites like this, and if so what are they. Rather than answer this question you attacked the very notion as absurd, so my edit was an attempt to defend its... non-absurdity? You're right, though; what's left is barely a question anymore. I may take one last (concise) crack at it yet.
Comment: @Dan: I did not "attack" the notion as absurd.  I pointed out that the notion is a contradiction.  "Open" == "Innovative".  "Single Best" or "standard" is the diametric opposite of innovative.  I'm suggesting that there cannot possibly be an answer to the original question that you would find satisfying because your have opposing criteria.
Answer: Here is something, maybe not the best implementations. But a book called Design Patterns contains what is considered by many programmers some of the best patterns to follow!


Unable to Install SQLEXPR_x64_ENU 2012 on windows 8 (vc_red.msi)

Question: This error occurs every time when I try to install SQL Server 2012 on Windows 8 64bit,The Feature you are trying to use is on a network resource that is unavailable
SQL Server Setup has encountered an error when running a Windows Installer file.
Windows Installer error message: The system cannot find the file specified.
Windows Installer file: d:\82c2346193c40f69152eb7\redist\VisualStudioShell\VC10SP1\vc_red.msi
Answer: you can tack a look here and it works for me
Unable to Install SQL Server Developer 2012 on windows 7 (vc_red.msi)
Answer: I had a similar issue on Windows 7 and the error message stumped me for a while.
Eventually I realised that the message was referring to my install of SQL 2008, and that the temporary location used during the install of 2008 was what the error was referring to. I guess 2012 needed to uninstall some 2008 bits and pieces.
In the end I rebuilt the temporary structure from my 2008 install and the installation continued successfully. Painful...


Calculus of variations -- how does it make sense to vary the position and the velocity independently?

Question: In the calculus of variations, particularly Lagrangian mechanics, people often say we vary the position and the velocity independently. But velocity is the derivative of position, so how can you treat them as independent variables?
Comment: Superb question aksink about the very fundament of everything we calculate. Furthermore - provoking great answers. You are very welcome to share your doubts with us @grizzly adam :) Greets
Comment: I have worried about this for years, it stopped me dead when trying to learn applied mathematics and I have met really good pure mathematicians who were similarly troubled.
One explanation which makes some sense to me is in the cheap book "Classical Mechanics - the theoretical minimum" which uses a school-like, infinitesimal, approach, to my mind answering a question the author does not actually pose.
Thanks for posting this question.
Comment: I recommend you take a look at Burke's book [*Applied Differential Geometry*](https://doi.org/10.1017/CBO9781139171786)  (Cambridge U. Press 1987). Consider that its dedication in the front pages says this: "*To all those who, like me, have wondered how in hell you can change $q$ without changing $\dot{q}$.*" :)
Comment: Related questions on Math.SE: https://math.stackexchange.com/q/1798396/11127 , https://math.stackexchange.com/q/580858/11127
Comment: Related: https://physics.stackexchange.com/q/60706/ , https://physics.stackexchange.com/q/119992/
Answer: Unlike your question suggests, it is not true that velocity is varied independently of position. A variation of position $q \mapsto q + \delta q$ induces a variation of velocity $\partial_t q \mapsto \partial_t q + \partial_t (\delta q)$ as you would expect.
The only thing that may seem strange is that $q$ and $\partial_t q$ are treated as independent variables of the Lagrangian $L(q,\partial_t q)$. But this is not surprising; after all, if you ask "what is the kinetic energy of a particle?", then it is not enough to know the position of the particle, you also have to know its velocity to answer that question.
Put differently, you can choose position and velocity independently as initial conditions, that's why the Lagrangian function treats them as independent; but the calculus of variation does not vary them independently, a variation in position induces a fitting variation in velocity.
Comment: To be more precise: it is not only about having to choose independent initial conditions. Velocities and positions as coordinates are always independent *unless* we are on a solution of the equation of motion. That is, $v^j=\dot{q}^j$ only on the trajectories that solve the Euler-Lagrangian equations. Onto those, variations of the former imply variations of the latter. Elsewhere they are unrelated.
Comment: @Theoretical In principle, the Lagrangian may depend on acceleration as well, e.g. be represented by a function $L(q,v,a)$ where $q,v,a$ are independent variables. However, it is only evaluated on curves $q(t)$ where $q\equiv q(t)$, $v \equiv \partial_t q(t)$ and $a \equiv \partial^2_t q(t)$.
Comment: I'm somewhat convinced by the explanation by @gented. But yet I cant help wringing my hands in frustration. I've gone thru Goldstein etc, they just gloss over this. Surely this conceptual leap of thought deserved ample discussion in the text. Fine, even if we go with what gented says (in other words, $q$ and $qdot$ are dependent albeit only when on the solution path), what still bothers me is this: Doesn't it yet have to be consistent? As in, after solving for the path, normally if we retrace our Math steps then we'd normally expect every step to be satisfied. Doesn't seem to be the case here.
Comment: For example, if $f = q_1^2t^3 + q_2^2t^2$. Suppose $q_1$ and $q_2$ are functions of time (that are unknown apriori but get known after solving the problem). Then, is it correct to say that $\frac{\partial(f)}{\partial(q_1)} = 2q_1t^3$, as all the justifications heretofore would have one believe? Well, that doesn't seem to be right if the solution was eventually found to be say $q_1(t)=t$. If we substitute this in $f$, we get $f=q_1^5 + q_1^2q_2^2$, and *then* calculate $\frac{\partial(f)}{\partial(q_1)}$, that gives a totally different value. How does one explain this?
Comment: Please explain the first few lines. Position and velocity are independent. They depend explicitly only on time. They off course depend implicitly on each other but not at all explicitly. You van not change v by jus changing x alone. When you change x it is understood that t changes. It is because of that change in t that v changes. Essentially the partiality derivative of v with x is 0 but the derivative of v with x is not 0. That is why, I think we don't apply any " chain rule " here !
Comment: @vharihar In this, case the issue is actually with the partial derivative — it is not stable under a change of variables! Consider for example $g(x,y)$. The partial derivative $\partial g/\partial x_{y=\text{const}}$ is different from the partial derivative $\partial g/\partial x_{z=\text{const}}$ where $z=y-x$.
Comment: @Shashaank  the derivative of v with respect to x **is** 0.
Comment: @Greg Graviton So according to your answer, if I can choose acceleration independently, then will it also be treated as an independent variable?
Comment: I have gone through the answer and the comments but again I wouldn't say I'm entirely convinced. I know that whether you're at (1,1) or (100,100), a velocity of 5i+2j remains 5i+2j.  This may not be the case for the same points in,say,spherical polar coordinates. However I am not being able dismiss the following idea: $x=t$ and $v=t^2$ which means $v=x^2$ or $\frac{df}{dx} \neq 0$. Can you give a convincing argument against this?
Comment: @AmbicaGovind You need to give a more well-defined statement to argue for or against. You have not defined $f$, do you mean $v$? Could you rewrite `\frac{dv}{dx}` in terms of its definition as limit, with particular attention to which of the variables $t$ or $x$ varies, and which does not?
Comment: @gented is incorrect. You can have paths where $v = \dot q$ that do not solve the Euler Lagrange equations that are just not stationary with respect to the action. Greg stated it correctly.
Answer: The answer to your main question is already given -- you do not vary coordiante and speed independently. But it seems that your main problem is about using coordinate and speed as independent variables.
Let me refer to this great book:
"Applied Differential Geometry". By William L. Burke. The very first line of the book (where an author usually says to whom this book is devoted) is this:

It is true that from time to time student do ask this question. But attempts to explain it "top down" are usually just lead to more and more questions. One really needs to make mathematical "bottom up" order in the topic. Well, as the name of the book suggest -- the mathematical discipline one needs is differential geometry.
I cannot retell all the details, but briefly it looks like this:

You start with a configuration space $M$ of your system. $M$ is a (differentiable) manifold, and $q$ are the coordinates on this manifold.
Then there is a specific procedure, that allows you to add all the possible "speeds" at every given point of $M$. And you arrive at the tangent bundle $TM$, which is a manifold too, and ($q$,$\dot{q}$) are different coordinates on it.
Lagrangian is a function on $TM$.
Comment: I have this book and tried to read it. But it lacks clear definitions, and I found it to be more frustrating than enlightening. Additionally, I don't believe that it is necessary to know differential geometry to understand calculus of variations. That's like saying that you can't understand arithmetic unless you know set theory.
Comment: I think to really appreciate Lagrangian and Hamiltonian mechanics you **do** need to understand some differential geometry. Arnold says in his book *Mathematical Methods of Classical Mechanics* that "Hamiltonian mechanics cannot be understood without differential forms." This book, by the way, will teach you the differential geometry you need to get started, assuming just some calculus.
Comment: First of all, as I said, you are mixing up two different points: about variational calculs and about independence of speeds and coordinates. Second -- I didn't say that you have to read only one book to understand DG.
Answer: Considering what Greg Graviton wrote, I'll write out the derivation and see if I can make sense of it.
$$ S = \int_{t_1}^{t_2} L(q, \dot q, t)\, \mathrm{d}t $$
where S is the action and L the Lagrangian. We vary the path and find the extremum of the action:
$$ \delta S = \int_{t_1}^{t_2} \left({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}\delta \dot q\right) \,\mathrm{d}t = 0\,. $$
Here, q and $\dot q$ are varied independently. But then in the next step we use this identity,
$$ \delta \dot q = {\mathrm{d} \over \mathrm{d}t} \delta q. $$
And here is where the relationship between q and $\dot q$ enters the picture. I think that what is happening here is that q and $\dot q$ are treated as independent initially, but then the independence is removed by the identity.
$$ \delta S = \int_{t_1}^{t_2} \left({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}{d \over \mathrm{d}t} \delta q\right) \,\mathrm{d}t = 0 $$
And then follows the rest of the derivation. We integrate the second term by parts:
$$ \delta S = \left[ {\partial L \over \partial \dot q}\delta q\right]_{t_1}^{t_2} + \int_{t_1}^{t_2} \left({\partial L \over \partial q} - {d \over dt}{\partial L \over \partial \dot q}\right)\delta q\, \mathrm{d}t = 0\,, $$
and the bracketed expression is zero because the endpoints are held fixed. And then we can pull out the Euler-Lagrange equation:
$$ {\partial L \over \partial q} - {\mathrm{d} \over \mathrm{d}t}{\partial L \over \partial \dot q} = 0\,. $$
Now it makes more sense to me. You start by treating the variables as independent but then remove the independence by imposing a condition during the derivation.
I think that makes sense. I expect in general other problems can be treated the same way.
(I copied the above equations from Mechanics by Landau and Lifshitz.)
Comment: "Not a word of Landau, not a thought of Lifshitz."
Comment: The notation for the arguments $L$ is somewhat confusing, in which case it is instructive to consider the following example: take $F(x,2x-y)$ and vary $F(x+\delta x,2(x+\delta x)-y) = \frac{\partial F}{\partial x}\delta x + \frac{\partial F}{\partial (2x-y)}2\delta x$. You might say that the arguments of $F$ are varied independently, but that sounds weird. If anything, it's just that the notation for the partial derivatives of $F$ is bad; it is much better to write $F(u,v)$ and $(u,v)=(x,2x-y)$ to obtain $\delta F = \frac{\partial F}{\partial u}\delta u + \frac{\partial F}{\partial v}\delta v$
Comment: ... and to express the variations $\delta u$ and $\delta v$ in terms of $\delta x$ afterwards.
Comment: Well, instead of saying "$q$ and $\dot q$ are varied independently", you could also say "$q$ and $\dot q$ are varied (perhaps independently, perhaps not)" and later note that the variation $\delta\dot q$ is given by $\delta\dot q = \frac d{dt} \delta q$.
Comment: Yes, the notation is confusing. That's another problem.
Comment: Landau is a great mathematical physicist but he's not known as a simple writer :-)
Comment: @grizzlyadam so in the end, they are not independent but can be treated as much because the mathematics prove it, right?
Answer: Here is my answer, which is basically an expanded version of Greg Graviton's answer.
The question of why one can treat position and velocity as independent variables arises in the definition of the Lagrangian $L$ itself, before one uses equation of motion, and before one thinks about varying the action $S:=\int_{t_i}^{t_f}\mathrm{d}t \ L$, and has therefore nothing to do with calculus of variation.
I) On one hand, let us first consider the role of the Lagrangian.
Let there be given an arbitrary but fixed instant of time $t_0\in [t_i,t_f]$. The (instantaneous) Lagrangian $L(q(t_0),v(t_0),t_0)$ is a function of both the instantaneous position $q(t_0)$ and the instantaneous velocity $v(t_0)$ at the instant $t_0$.
Here $q(t_0)$ and $v(t_0)$ are independent variables. Note that the (instantaneous) Lagrangian $L(q(t_0),v(t_0),t_0)$ does not depend on the past $t<t_0$ nor the future $t>t_0$. (One may object that the velocity profile $\dot{q}\equiv\frac{\mathrm{d}q}{\mathrm{d}t}:[t_i,t_f]\to\mathbb{R}$ is the derivative of the position profile $q:[t_i,t_f]\to\mathbb{R}$, so how can $q(t_0)$ and $v(t_0)$ be truly independent variables? The point is that since the equation of motion is of 2nd order, one is still entitled to make 2 independent choices of initial conditions: 1 initial position and 1 initial velocity.) We can repeat this argument for any other instant $t_0\in[t_i,t_f]\,.$
II) On the other hand, let us consider calculus of variation.
The action functional
$$S[q] ~:=~ \int_{t_i}^{t_f}\mathrm{d}t \ L(q(t),\dot{q}(t),t)\tag{1}$$
depends on the whole (perhaps virtual) path $q:[t_i,t_f]\to\mathbb{R}$. Here the time derivative $\dot{q}\equiv\frac{\mathrm{d}q}{\mathrm{d}t}$ does depend on the function $q:[t_i,t_f]\to \mathbb{R}\,.$ Extremizing the action functional
$$\begin{align}0~=~&\delta S \cr
~=~& \int_{t_i}^{t_f}\mathrm{d}t\left[\left.\frac{\partial L(q(t),v(t),t)}{\partial q(t)}\right|_{v(t)=\dot{q}(t)} \delta q(t) \right.\cr
&+\left.\left.\frac{\partial L(q(t),v(t),t)}{\partial v(t)}\right|_{v(t)=\dot{q}(t)}\delta \dot{q}(t)\right]\cr
 ~=~& \int_{t_i}^{t_f}\mathrm{d}t\left[\left.\frac{\partial L(q(t),v(t),t)}{\partial q(t)}\right|_{v(t)=\dot{q}(t)} \delta q(t) \right.\cr
&+\left.\left.\frac{\partial L(q(t),v(t),t)}{\partial v(t)}\right|_{v(t)=\dot{q}(t)}\frac{\mathrm d}{\mathrm{d}t}\delta q(t)\right]\cr
~=~& \int_{t_i}^{t_f}\mathrm{d}t\left[\left.\frac{\partial L(q(t),v(t),t)}{\partial q(t)}\right|_{v(t)=\dot{q}(t)} \right.\cr
&-\left. \frac{\mathrm d}{\mathrm{d}t}\left(\left.\frac{\partial L(q(t),v(t),t)}{\partial v(t)}\right|_{v(t)=\dot{q}(t)} \right)\right]\delta q(t)\cr
&+ \int_{t_i}^{t_f}\mathrm{d}t\frac{\mathrm{d}}{\mathrm{d}t}\left[\left.\frac{\partial L(q(t),v(t),t)}{\partial v(t)}\right|_{v(t)=\dot{q}(t)}\delta q(t)\right] \end{align}\tag{2} $$
with appropriate boundary conditions leads to Euler-Lagrange (EL) equation, which is the equation of motion (EOM).
$$
\begin{align} \frac{\mathrm d}{\mathrm{d}t}&\left(\left.\frac{\partial L(q(t),v(t),t)}{\partial v(t)}
\right|_{v(t)=\dot{q}(t)} \right)\cr
&~=~ \left.\frac{\partial L(q(t),v(t),t)}{\partial q(t)}
\right|_{v(t)=\dot{q}(t)} ~.\tag{3}\end{align}
$$
III) Note that
$$\frac{\mathrm{d}}{\mathrm{d}t}~=~\dot{v}(t)\frac{\partial}{\partial v(t)}+\dot{q}(t)\frac{\partial}{\partial q(t)}+\frac{\partial}{\partial t} \tag{4}$$
is a total time derivative, not an explicit time derivative $\frac{\partial}{\partial t}$, so that the EL equation (3) is really a 2nd-order ordinary differential equation (ODE),
$$\begin{align} \left(\ddot{q}(t)\frac{\partial}{\partial v(t)}
+\dot{q}(t)\frac{\partial}{\partial q(t)}
\right.&\left.
+\frac{\partial}{\partial t}\right) \left. \frac{\partial L(q(t),v(t),t)}{\partial v(t)}
\right|_{v(t)=\dot{q}(t)} \cr
~=~& \left.\frac{\partial L(q(t),v(t),t)}{\partial q(t)}
\right|_{v(t)=\dot{q}(t)}~. \tag{5}\end{align}$$
To solve for the path $q:[t_i,t_f]\to \mathbb{R}$, one should specify two initial conditions, e.g.,
$$q(t_i)~=~q_i\qquad\text{and} \qquad\dot{q}(t_i)~=~v_i.\tag{6}$$
Comment: There might be a better answer in the same line with this post and Christian Blatter's comment https://math.stackexchange.com/a/1211868/603316. The "derivative" between the two function wasn't really derivative of the variables, rather viewed as functional derivative, $\frac{d q(t_2)}{d \dot{q}(t_1)} = \frac{d \int_{\tau}^{t_2} \dot{q}(t) dt}{d \dot{q}(t_1)} =0$ and this was a functional variation.
There might be issue with the contact terms, $\delta(t_1-t_2)$, which was in the region of qm.
Comment: Hi. "We can repeat this argument for any other instant" - I get your point only for the initial conditions. At initial conditions, velocity and position are definitely independent, but not at any other point of time during the path. Can you elaborate ?
Comment: Hi @Chemistry. Thanks for the feedback. There are 2 independent conditions. These 2 can be imposed at any instant $t_0$. They don't have to be at the initial instant.
Comment: @Qmechanic Imagine L being $\frac{1}{2}m\dot q(t)^2 - mgq(t)$ where $q(t)$ and $\dot q(t)$ are functions of t. To know L's value, let's say at $t=2$, you would plug in $t=2$ and get L. This means that behind the hood, the speed that it calculated is dependent on the position,  but note that since $q(t)$ is the path which satisfies the initial conditions at $t=2$, $q(t)$ and $\dot q(t)$ would in turn return the values of that initial conditions for $t=2$, which we can say that at that instant($t=2$), position and speed are still independent. Whatever $t$ you choose, the same logic applies.
Answer: While it is true that the function $\dot{q}(t)$ is the derivative of the function $q(t)$ w.r.t. time, it is not true that the value $\dot{q}$ is at all related to the value $q$ at a given point in time, since a value is just a number, not a function.  The action is a functional of $q(t)$, and so it would make no sense to vary the action both w.r.t. $q$ and $\dot{q}$.  But the
Lagrangian $L(q,\dot{q})$ is a function of the values $q$ and $\dot{q}$, not a functional of the functions $q(t)$ and $\dot{q}(t)$.  We can promote $L$ to a function of time if we plug in $q(t)$ and $\dot{q}(t)$ instead of just $q$ and $\dot{q}$.  (Remember a functional turns a function into a number, e.g., $S[q]$, whereas a function turns a value into a number, e.g., $L(q,\dot{q})$.
To solve for $q(t)$ we extremize the action $S$, by demanding that it is extremal at every point, $t$.  This is equivalent to solving the Euler-Lagrange equations at every point $t$.
Since at any point $t$ the values $q$ and $\dot{q}$ are independent, they can be varied independently.
Answer: If we have a function $f(x,v)$, the partial derivatives are defined by $$\frac{\partial f(x,v)}{\partial x} \equiv \lim_{h\to 0} \frac{f(x+h,v)-f(x,v)}{h}$$ and $$\frac{\partial f(x,v)}{\partial v} \equiv \lim_{h\to 0} \frac{f(x,v+h)-f(x,v)}{h}$$ This implies, for example, for $f=v^2$ that $$\frac{\partial v^2}{\partial x} \equiv \lim_{h\to 0} \frac{v^2-v^2}{h} =0.$$ Moreover, for $v= \frac{dx}{dt}$ we find that $x \to x+h$ implies $v = \frac{dx}{dt} \to v' = \frac{d(x+h)}{dt} = \frac{dx}{dt} =v$. Thus $$\frac{\partial \frac{dx}{dt}^2}{\partial x} \equiv \lim_{h\to 0} \frac{\frac{dx}{dt}^2-\frac{dx}{dt}^2}{h} =0.$$
Hence it makes sense to consider the partial derivatives of the Lagrangian with respect to $x$ and $v$ separately and in this sense treat them independently.

In more physical terms, recall that our goal in the Lagrangian formalism is  to figure out the correct path in configuration space between two fixed location. A path is characterized by a location and velocity at each point in time. We are as general as possible and consider really all possible paths. This implies that we consider all possible pairings of locations and velocities. The physical classical path is special for two reasons:

it's a solution of the Euler-Lagrange equation (= extremum of the action)
the locations and velocities at each moment in time are related by $v \equiv \frac{dq}{dt}$. (If you want, $v \equiv \frac{dq}{dt}$ is the second equation that we need in the Lagrangian formalism analogous to how there are two Hamilton equations in the Hamiltonian formalism. The second Hamilton equation defines the canonical momentum as a derivative of the Lagrangian. For general paths in phase space, any pairing of location and momentum is possible. Only for the physical classical path we find canonical momentum values that are given as the appropriate derivative of the Lagrangian.)
Comment: The definition of the partial derivative is a mathematically convincing reason in a way. But I still don't understand how $\frac{\partial v}{\partial x}$ can be zero if, for example $v=t^2$ and $x=t$ in which case $v$ can be written as $x^2$
Comment: Because by definition $\frac{\partial v}{\partial x}$ means the derivative of $v$ w.r.t. $x$ with $v$ kept constant. If you impose $v = x^2$ then $x^2$ must be kept constant (the appropriate variables are now $x$ and $x^2$) and the derivative is still zero.
Answer: The derivative of a function $f(t)$ is the function $\dot{f}(t)$ in general different than $f$, and in the general case the two are not even linearly dependent, which is simple to see if you take the Taylor expansion. It is only after you define differential equations with them that they are linked algebraically, and this is what the calculus of variations does.


Windows 7 interfering with smart cards

Question: I have an application that uses the PC/SC API to communicate with smart cards.  On Windows 7, I get strange results: the data returned from the cards is inconsistent and invalid with certain commands.  If I disable the Smart Card Plug and Play service in group policy then everything works fine.  Has anyone experienced anything similar?  Is there any way to get the smart card plug and play service to play nice?  It would be nice to not have to disable it...
Comment: See also:
http://www.lazybit.com/index.php/2010/10/20/smart-card-driver-for-windows-7-popup?blog=2
Answer: This is happening because you are using a smart card that supports "plug-and-play" under windows 7. In this situation, Windows 7 will automatically attempt to download and install a "Card Mini Driver" for the card when it is inserted. It sounds like the Card Mini Driver that is getting installed is causing the inconsistent behavior.
The obvious workaround is to use a card that doesn't support "plug-and-play" under windows 7. If this isn't possible, you will need to post additional details about the problems you are seeing, specifically what kind of smart card are you using, and what commands are breaking after the card mini driver is installed?
Technet details on Smart Card Plug and Play in windows 7 can be found here:
http://technet.microsoft.com/en-us/library/dd979547(WS.10).aspx

Windows 7 interfering with smart cards


jsonReader for this JSON schema

Question: I am trying to get a jQGrid related to a file stored in the server which look similar to this one:
    {"newRow":[{
                 "rowId":"r02",
                  "address": [{
                                  "pc":"02020",
                                  "city":"Madrid"
                            ]},
              ]},
     "newCell":[{
                   "rowId":"r02",
                   "cellId":"r02",
                   "outCome":"200"
               ]},
     "newCell":[{
                [{
                   "rowId":"r02",
                   "cellId":"r03",
                   "outCome":"220"
               ]}
               ]},
     "newProvider":[{
                      "cellId":"r03",
                      "name":"Peter"
                   ]},
     "newProvider":[{
                      "cellId":"r03",
                      "name":"Jay"
                   ]},
     }

For what I read in some links, it should have another structure but I also read that is possible to modify the jsonReader to adapt it to my structure. But I have no idea of how to do this. Thanks in advance.
Answer: You have to use jsonReader. Here is a linf to the documentation.
You should use it as a function. It looks kie that.
jsonReader: {
  repeatitems: false,
  id: "Id",
  root: function (obj) { return obj; },
  page: function (obj) { return 1; },
  total: function (obj) { return 1; },
  records: function (obj) { return obj.length; }
}

jsonReader for this JSON schema


name of last person who modified node - Display suite

Question: with Display suite in Drupal 7 I can get the last modified date , but how can I display which user did the last modificatioin?
Answer: The easiest way to see who did the last revision is to, well, enable revisions for that content type and require them. This won't help you with past edits, but going forward, every edit will create a new revision, which will have a user associated with it, so you can see who made the last edits.
Comment: In the view I get a choice to see revision: updated date, and log message, but not the user who did the last update.  How do I see that in the view?
Comment: answer of how to see it here http://drupal.stackexchange.com/questions/178402/how-to-see-node-revision-user-in-a-view


Why do rockets not fly straight up after launching; wouldn't that be less atmosphere to traverse?

Question: If you look at any launch it appears as though rockets do not take the shortest path. Their trajectory bends. Why do rockets not fly straight up after launching; wouldn't that be less atmosphere to traverse?
Comment: They are going into orbit around the equator, so It's not like they can go straight up and do a 90.
Answer: Since orbiting isn't about altitude, but is instead about velocity (orbiting is the art of throwing yourself at the ground, and missing because you're going sideways so fast), the final goal is to have a heap of lateral velocity so that gravity is merely curving your course, instead of actually pulling you to the ground.
The most efficient flight to orbit is essentially a nearly-horizontal trajectory over the horizon, only gaining what altitude you need to avoid collisions.  Any additional flight up whatsoever is a concession to the atmosphere and should be given up as quickly as possible.
There's a number of forces that an ascending rocket has to overcome, atmospheric drag is merely one of them.  Gravity is also a huge drain that needs to be overcome by any ascending vehicle. Flying straight up means that 100% of the energy put into the rocket's velocity is available to be extracted by gravity.
By flying even slightly sideways, all of the lateral motion generated is energetic money in the bank, virtually immune to losses from gravity, and only subject to losses to atmospheric drag.
After an initial ascent, the air density has dropped rapidly and further benefits from altitude begin to taper off:

Depending upon the aerodynamic performance of the rocket's airframe, between 5-10km up, you've gotten rid of the lion's share of atmospheric drag, but still have basically all the same gravity to contend with.
These two forces are managed in a balance, so the more the rocket ascends, the more beneficial it is to turn sideways and get more of the energy from your motor into your lateral velocity.
The good news is that as you add lateral velocity, you also get less atmospheric drag by turning further into that lateral direction, gravity is reducing your 'up' meaning the ratio of side-to-up is more and more weighted to the side; this is also your angle of attack against the air.
Answer: Yes, but atmospheric drag is nearly negligable.
To get into orbit, the rocket has to expend energy doing several different things, and they each have performance constraints that affect the structural weight of the rocket.
Roughly in order of fuel consumption, the rocket has to

Gain orbital velocity (a reversible kinetic energy term)
Overcome gravity loss (a non-reversible term for lost impulse)
Gain orbital altitude (a reversible potential energy term)
Overcome atmospheric friction without getting too hot, overstressing the structure, or sliding sideways through the air (because that increases the drag).

If you look at only the first and third, they are both reversible and you can run any linear combination, including launching straight up and then kicking over 90 degrees and gaining transverse velocity. (Essentially, all paths would have the same fuel requirement.)
The trouble with doing that is that the gravity loss is path dependent. If a rocket was hovering just above the ground, it is still burning fuel, but not gaining kinetic energy and not gaining potential energy. The rocket has to overcome all of local gravity to stay in the air. But if we begin to traverse, the ground is falling away from us and we only need to overcome, say 90% of gravity and let the rocket accelerate towards the ground which it will never hit. The total gravity loss is the time integral of gravity force - centrifugal force. The ideal solution (with gravity, but without atmosphere) is a very high acceleration rate and short burn time, followed by a coast into orbit. Then you need a second burn near apogee to not crash back down. If you can accelerate hard enough, you can just launch sideways. But we can't accelerate that hard.
The atmosphere puts all kinds of practical constraints on the coupled problem of vehicle design and launch trajectory design. The way it shakes out, the atmospheric drag ends up being nearly negligible. Here are some Delta V numbers from Stanford course material (Delta V is the change in velocity, where all of the rocket engine's efforts have been converted to what would happen in space.)

So even with a well executed launch, the atmospheric drag loss is only 5% of the gravity loss and only 1% of the total Delta V budget. A poor trajectory can increase the gravity loss by more than the total atmospheric drag. The trajectory is driven largely by attempting to minimize the gravity loss figure. The atmosphere's real nuisance value is the structural loads that are very peaky around mach 1.4. So we have to go kind of slow until the air is thin enough to be able to push though it with a light upper structure.
Answer: Apart from the intentional trajectory adjustments to obtain the right orbit there is still another another reason, that would the rocket would appear to curve.
Even  if you a rocket is shot in a straight line from anywhere in the world (apart from the poles), you would see it curving, if you were an observer stationed on earth.
The reason for that is the Coriolis Effect. The Coriolis acceleration is:
$$a_C= -2\omega \times v $$
where:

$\omega$ is the rotational velocity of the earth ($7.29 \cdot 10^{-5} radians/second$).
$v$ is the velocity of the rocket.

For an observer on a rotating frame (see earth), the linear motion of a body in an inertial frame of reference would appear to be curving.
A nice video, showing the Coriolis effect on a rotational frame can be found in this Youtube link (imagine the ball is the rocket).
Normally, the Coriolis acceleration is quite small due to the small value of $\omega$, however because the velocity of a rocket is quite high in a direction perpendicular to the angular velocity of the earth, the effect quickly starts to become noticeable and more pronounced as the velocity builds up.

This is an excerpt from a NASA exercise paper.

The Coriolis effect is the name given to the imaginary force that deflects objects, such as rockets or large storms, which move over the surface of some
planets.

Also have a read at this: why is it better to launch rockets from the equator

When a spacecraft is launched into orbit, it should end up spinning around the Earth quickly enough not to be pulled back in by the Earth's gravity. The huge rockets used in launching a spaceship help this to happen by giving a huge amount of thrust, enough to achieve escape velocity. However, the spin of the Earth itself can help give it a push as well. Anything on the surface of the Earth at the equator is already moving at 1670 kilometers per hour.

Why do rockets not fly straight up after launching; wouldn’t that be less atmosphere to traverse?


How do people at service center check if the phone is rooted or not?

Question: How do they check it, in order to decide whether the phone is in warranty or not?
I am referring to the majority of the phone manufacturers who void the warranty of their phones 'as soon as they are rooted'.
Answer: Varies as per OEM

Most devices need bootloader to be unlocked before rooting. OEMs like Motorola insist on a formal request to be made on their website for obtaining boot loader unlock code. The moment you make the request, your warranty is void whether you unlocked the bootloader /  rooted or not

Edit: Since you own a Motorola device, more details

When you request for bootloader unlock code, you have to identify your device serial number, which means database is available with service center to check against your serial number. Motorola clearly warns you that requesting for unlock code  = rooting, whether you root or not and void warranty

If you do unlock bootloader and do not take steps to prevent this warning from showing up (involves flashing a custom boot logo file) , this image will show up everytime you boot.

Even if you lock bootloader again by applying a bootloader lock code, in bootloader mode it will show
`DEVICE IS LOCKED. STATUS CODE : 2'

whereas the status code for unlocked devices is 1
So be very sure about unlocking bootloader, the service center guys can easily find out !!

For Samsung devices it is very easy to read the Knox status. If that is tripped, your warranty is void ( exceptions apply in some geographies)
Some devices display Modified / Custom status in the device properties or in diagnostics ( this to Andy for pointing that out)
It is reasonable to expect that OEMs who void warranty on rooting would have a root status shown in the device diagnostics. Just in case there isn't, they can  use Root Checker apps, or check using Terminal Emulator. Use of third party apps I guess would be more of an exception, since inbuilt tools would do the job
Comment: Also, on Samsung, even back in the days where Knox does not exist, there's still a flag for modified /system (that says OFFICIAL or CUSTOM).
Answer: Install Root Checker app or connect it to computer and check its firmware.
You can connect it to computer through Mobogenie and check it's root status
Comment: The question is not asking how a user can check, but how the oem will check.


How can the speed of light change in the medium when we know that it is always equal to $c$?

Question: How can the speed of light change in the medium when we know that it is always equal to $c$? If we say the speed of light is changing in the medium, it will contradict the Einstein's law of special relativity.
Comment: Related: http://physics.stackexchange.com/q/11820/2451 , http://physics.stackexchange.com/q/466/2451 and links therein.
Comment: I didn't know I knew that! ;-) But it's not true: light always travels at $c$ in vacuum but in all other media its velocity $<c$.
Answer: James and Griffiths, Am J Phys 60, 309-313 1992, treat the transmission and reflection of a plane em wave normally incident on a transparent medium. Using a perturbative approach they argue that the incident electric field polarises the medium, the oscillating dipoles associated with this polarisation radiate gives rise to an additional electric field, which in turn gives rise to an additional polarisation ..... the "radiation from many induced molecular dipoles conspires to produce a single wave propagating at the reduced speed".
Comment: This is the classical physical model; Fresnel's waves, plus the physics of Maxwell's equations. It does not explain (or even use) photons.
Comment: Sorry, yes you are correct, this is the classical model.
Answer: The mean speed changes, because the photons scatter off the atoms in the medium which makes their path longer. However, the momentary speed is always equal to $c$.
Comment: As this answer comes from Photon, it surely is first hand information.
Comment: This is a very misleading model. For example, it suggests that the photons are randomly changing direction as they bounce onto atoms (which would make them quickly forget their original direction), while in reality they propagate in a straight line. It also doesn't explain the case $n < 1$, which has been observed.
Comment: When you say scatter would that be like wiggling back and forth between the atoms as it crosses the medium? I have always wondered if the time and a path like that have ever been compared? Is there an experiment somewhere that has been conducted?
Comment: The form of scattering within the media must preserve the optical coherence, or else the image is lost -- and the media is no longer transparent!
Comment: @knzhou: You are totally right, thanks for your comment
Answer: Transparent materials (glass, air) transmit images; if the image is distorted or indistinct, we know that the material is altering the coherence of the optical information. That is, what started out at the beginning has not arrived all at the same time.  With enough distortion the image is completely lost.
So what is required for a transparent medium to successfully transmit an image? Since light is a physical wave, the transparent medium must preserve the coherence of the phase information of the light.  In a typical glass the phase front is slightly slowed while traveling through the glass; this slowing is encoded in the index of refraction, $n = c/v$.
If the material absorbs some frequencies, the material will appear to be colored; a photon that is absorbed (depending on the energy level structure) can be re-emitted, but this will be at (a) a random time later, and (b) in a random direction. No image for this color!  There is an exception: stimulated emission, which is the key to building a laser. But this is not how images are transmitted in a passive material.
The process that transmits images can be summed up as Coherent Forward Scattering: Coherent, because otherwise the image integrity is reduced; Forward, because the image is transmitted in this direction, through the material; and Scattering, the remaining available generalized mechanism at the quantum level.
The result is quite like the Huyghen's wavelet model for light transmission: the photons are the waves that are scattered coherently, and because it is coherent, they are able to interfere both constructively and destructively to maintain the coherence of the overall phase front.

The wavefront slows in the medium due to the coherent forward scattering of the photons; this slightly retards the wavefront, even though the photons are still travelling at $c$
It is the interference that slows the phase velocity through through the material; the individual photons continue to "move" at the speed of light, $c$, but the effective motion of the phase front is slowed.
Richard Feynman devotes some time to this in his lectures on QED: The Strange Theory of Light and Matter
Borrowed in part from my answer to Why is everything not transparent?


Is it fardh on me to wear a niqab?

Question: I was 14 when my father and mother forced me to wear a niqab without even describing me the benefits of this. Obviously, I knew but I never had that sense of acknowledgement in my head. On their force and beating, I started wearing niqab just to avoid any conflict in the house. It's been more than 7 years, I've been wearing it but I never loved this act with my full heart. I always wanted to take it off. Now, I'm 22 and in a medical field. I wore abaya and niqab whenever my parents are around and took them off in medical school. I dress modestly and also in favour of complete hijab. But whenever I wear niqab it's like I'm being hypocrite or dual faced because I'm not ready for this yet. I've discussed with my parents and every time instead of understanding me they start telling me that you're a kafirah you'll go to hell on this. You'll take us into hell. That's wrong. My family is Islamic extremist, they force teachings on everyone. Will I go to hell? What's it like to live in a house where you've to follow Islam by force and beating. I feel suffocated while living with them. What does Islam say about parents who force their children.
Comment: Does this answer your question? [Can I be forced by my parents?](https://islam.stackexchange.com/questions/49746/can-i-be-forced-by-my-parents)
Comment: Also see https://islam.stackexchange.com/questions/80292/ and regarding obedience in niqab.
Answer: sorry to hear that, many madhab believe that it is not compulsory to wear "niqab" it is base on your preference it is (mustahab) and you know islam is not a religion that force. As long as you dress modest you should be fine but as you said, you do it to "avoid conflict" allah said in surah is "۞ وَقَضَىٰ رَبُّكَ أَلَّا تَعْبُدُوٓا۟ إِلَّآ إِيَّاهُ وَبِٱلْوَٰلِدَيْنِ إِحْسَـٰنًا ۚ إِمَّا يَبْلُغَنَّ عِندَكَ ٱلْكِبَرَ أَحَدُهُمَآ أَوْ كِلَاهُمَا فَلَا تَقُل لَّهُمَآ أُفٍّۢ وَلَا تَنْهَرْهُمَا وَقُل لَّهُمَا قَوْلًۭا كَرِيمًۭا ٢٣" For your Lord has decreed that you worship none but Him. And honour your parents. If one or both of them reach old age in your care, never say to them ˹even˺ 'ugh,' nor yell at them. Rather, address them respectfully" you are not Kafira and inshallah you will not go to hell, the first hijab Allah put was in this ayat "tell the believing men and women to lower their gaze and gurad their modesty" if people were doing that then we would not have this problem. Islam teaches that every individual has the right to choose their own path in life, and that no one should be forced to do something that goes against their conscience. The Prophet Muhammad (peace be upon him) said, "There is no compulsion in religion" keep doing du'a for them and teaching them, may allah have mercy on us.
Comment: "many madhab believe that it is not compulsory to wear "niqab" it is base on your preference it is (mustahab) and you know islam is not a religion that force" this statement is self-contradictory I assume that you meant to say that it is **not compulsory**, but the first part of your sentence says otherwise.
Answer: You can wear a niqab if there is chance of you being kidnapped by some groups who believe it is lawful to kidnap girls and rape them. This way you can conceal your beauty and idea of kidnapping won't come to them. Actually this niqab practice came to existence from this fear and you can see non Muslim groups in countries like India covering their face. If there is no chance of kidnapping like in modern societies, it is better not to cover the face as it is your identity that Allah has given you.  Imagine if Allah has made all our faces same, how we will be able to identify mutually. So if everyone wear niqab, how can people identify each other. So if you are in medical field, you should not wear niqab as your identity is important in the field. Prophet (s.a) forbade covering the face even during circumambulating the ka'ba when people of both sex come very close to each other. If people can uncover face in the most sacred place, there is no problem in uncovering face elsewhere if no harm is feared.
About parents who force you, you can follow then in good matters only (معروف) and there is no duty on you to obey them in matters that are against religion. You can argue with them in good way and convince them that it is not required in Islam.

Is it fardh on me to wear a niqab?


How to prevent cross site scripting in wordpress?

Question: I did source analyzer with third party services to my wordpress project. It found out that there are heaps of cross site scripting vulnerabilities. One of the example is like image below. How do I prevent the cross site scripting? Should I put validation in front end view (HTML / CSS / JS) or on back end (PHP / Wordpress)?
Answer:
How do I prevent the cross site scripting?

Perform input validation and output encoding. Only allow characters that you expect, for example if you only expect digits, the back-end should not process regular characters.
Sometimes you expect a specific length, if the back-end detects it's either smaller or larger, do not process it.
Same goes for special characters, if you request a first name, don't allow special characters such as "<", ">", "$" etc.

Should I put validation in front end view (HTML / CSS / JS) or on back end (PHP / Wordpress)?

Validation from a security perspective is always performed server side. Client side validation is only done for user experience.
Answer: Use a strict content security policy (CSP).  See https://developer.mozilla.org/en-US/docs/Web/HTTP/CSP for more info.


Logging using Log4j taking long time

Question: I am using Log4j logging framework to insert the log into oracle database.But the insert query in the log4j properties file is taking a lot of time to execute and making the application very slow.When I removed the logging statements from the java code, the application worked fine.At first, I thought that the insertion into DataBase is taking time , but writing the log on an external file also takes a lot of time.
Can anyone please suggest a solution?
Thank You,
Dhaval Maheshwari.
Answer: If you application is under development then log level should be debug and before logging you should check for isDebugEnabled() and then log your string.
but If your application is in production then log level should be info and you must log minimal information in log file.

Always use atleast two log level in your application one for debuggnig
  mode(for development environment) and another for production mode and
  production log should be minimal.

This is the way you can speed up you applicaiton.

and second thing if you want to persist your logs into database then
  create a scheduler task whose responsibility would be reading logs
  from flat file and persisting them into database and schedule this to
  run only once in a day.
Answer: I suggest not to follow the technique u r following now.
First of all I am not sure why u r trying to log the output of log4j in DB.
Anyways if it is that necessary try something like this. Let the logfile write into a file as it is and later run a thread to dump this file from the disk when file is closed to the database as a batch process.
In this case your application will be separated from the latency of DB.
There are other solutions also using a JMS.
Where you can write it to a JMS queue and the consumer on the other hand can read the queue and write it a DB.
It depends on the kind of problem you are trying to solve though.
See of it helps
Answer: In Logging there are levels included in. For example in production only log application level exceptions and errors[ERROR level].
If it's tracking logs(Such as user actions) don't write them to files, directly add them to database. Hope this helps.

Logging using Log4j taking long time


What's the rectangle surrounding a font glyph, including its sidebearings, called?

Question: What's the rectangle surrounding a font glyph, including its sidebearings, called? I Googled the words "bounding box" and got conflicting results, where some sources say it's the rectangle that bound the glyph tightly (without sidebearings), and others say it includes sidebearings.
Comment: Not everything has a name unfortunately. There may in fact be no name, or multiple names for this.
Comment: In German it's "Schriftkegel"
Comment: @joojaa After hours of futile research, even going so far as looking into non-English sources, I reluctantly agree.
Comment: @JulianSteinmann you should convert that to an answer. THough ists not exactly the same thing, but close enough
Answer: It is possible that this does not have an official, authoritative name. There is certainly no need for this name in the font specification because this is a calculated value, not really needed for anything in the font drawing.
You're in luck! You can make the authoritative name.
Comment: I talked to a font designer and he said that it does not have a name but if it does its M size
Answer: Okay, so after doing quite a bit of research, here's my conclusion: there seems to be no perfect English term. But there is one workable candidate, although it won't be very useful in English: "body". This is a reference to the "body" of a sort in movable typesetting, as suggested by user K Pease on Typedrawers. It is in fact still used in Japanese type design, specifically as 仮想ボディ ("imaginary body"), given that "bodies" are much more important as a concept in East Asian calligraphy and typography than Western ones. "Imaginary body", although a literal translation, is indeed used in some very limited English materials translated from Japanese, so it's not completely terrible. Adobe translates this term as "embox", which can be confused with the concept of the "em square", but given that the height of this box is always exactly one em, it's not a bad one either. The only problems with "imaginary body" and "embox" is that I haven't found any concrete demonstration of these terms on Western text in Japanese sources (most examples are strictly about Japanese text), but I hazard it wouldn't be that different anyway.
In short, "body", "imaginary body" or "embox". Not perfect, but still usable to some extent.


What is the word "ô" ?

Question: I saw it in a historical info sign in the sentence "Ce métier ô combien difficile et périlleux a disparu progressivement avec le temps." I was astonished to encounter a one-letter word that I'd never seen before. The wiktionary entry gives no definition, and the TLF entry is a bit confusing. What does this word mean? Is it very formal, or very rare?
Comment: In fact the French wiktionary entry does give [a definition](https://fr.wiktionary.org/wiki/%C3%B4).
Comment: Same O as in "O Captain! My Captain!" - except with a hat.
Answer: The Latin word o that had a wide range of usage (From the Gaffiot: to call, invoke, wish, express surprise, indignation, happiness, pain, …), has produced three French interjections pronounced the same way: oh, ho, and ô. They have some overlapping meanings but also their own specificity.
The ô variant is more literary, that's likely the reason why you hadn't seen it before.
Unlike the others, ô is always followed by another word while oh and ho can be used alone or as part of a short exclamatory sentence. The one word sentence "ô ! " is "forbidden".

Oh ! combien de marins, combien de capitaines… - Victor Hugo
Ô rage ! ô désespoir ! ô vieillesse ennemie ! N'ai-je donc tant vécu que pour cette infamie ? - Corneille
Oh là la ! (Oh, boy!, Yikes!)
Ho ! Hisse ! (Heave-ho!, one, two, three, hup!)
Ho ! combien de fois il faut te le répéter ? (Hey!)
Ce pont, ô combien nécessaire, … (much needed)

While ô combien might sometimes be used an ironical way, there is absolutely no sign of irony in your quote:

Ce métier ô combien difficile et périlleux a progressivement disparu avec le temps
This occupation, how difficult and perilous was it !, has faded away over time.

Because of the exceptional tides and the shifting sands, the Moulières d'Honfleur were risking their lives collecting the mussels.
Comment: @LUNA *ô combien* is often but not always ironical. There is no sign of irony in the OP quote "*ô combien difficile et périlleux*" and no risk for it to be understood that way unless of course the context tells otherwise. This is definitely not the case after identifying the original source.
Comment: Ha "it is forbidden." One of the strictest and most important laws of France.
Comment: Very true, and perhaps the most important one!
Comment: @jlliagre As far as I understand now, an ironic tone is the key when you use this "ô combien + adjective" expression. As such, if you don't insinuate irony in any way whatsoever and just want to emphasise straightforwardly what you feel (for instance, "how beautiful!" with no sense of irony involved), should you use some other expression? Merci.
Answer: The word "ô" is essentially an outdated way of saying "oh !", and rarely used by itself except as part of this set phrase "ô combien + adjective" that means "(this job) oh so (difficult and dangerous)", expressing awe in an emphatic and often ironic fashion.
ps : J'ai trouvé ça drôle, moi aussi, la première fois que je l'ai rencontré.
Comment: Please note it is not outdated, but _literary_, and is seemingly not mostly used with _combien_ or any set phrase. See [this](http://larousse.fr/dictionnaires/francais/%C3%B4%C2%A0_/55329/difficulte), [this](http://bdl.oqlf.gouv.qc.ca/bdl/gabarit_bdl.asp?t1=1&id=3712) or [this](http://www.cnrtl.fr/definition/%C3%B4). Thanks.
Comment: Famous line from Le Cid: *Ô rage ! ô désespoir ! ô vieillesse ennemie !*.
Answer: From the Wiki: "O is used in the vocative, that is, as an invocation, in order to address something or someone."
"With how sad steps, O Moon, thou climb'st the skies!" -- Sir Philip Sydney
Comment: Hi Linda, this would benefit from examples in French.  As it stands, it seems to address the English.  Please see the [help] and [tour], et bienvenue sur French language SE.
Comment: As it's currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer).
Comment: This answer is correct. But you need to add a French source for the French version.

What is the word "ô" ?


CKEditor in CodeIgniter

Question: I wanna load CKEditor in CodeIgniter,I search a lot,but can't understand their way.
I placed ckeditor in application/plugins folder and now I wanna make editor ,so I do following in Controller Method.
include APPPATH.'plugins/ckeditor/ckeditor.php';
$CKEditor = new CKEditor();
$CKEditor->basePath = '/'.APPPATH.'plugins/ckeditor/';
$initialValue = '
This is some sample text.
';
echo $CKEditor->editor("editor1", $initialValue);

but it makes simple teaxaria only ,with This is some sample text. value.
where is the problem and how should I solve it?
Answer: I use this steps to add ckeditor to my codeigniter apps:
1) Download these files:

This for Ckeditor: http://pastebin.com/fkK9e0RR
This for Ckfinder: http://pastebin.com/SvyypmX4

2) Copy the files you just downloaded into your Application/libraries folder
3) Download the ckeditor helper here: http://pastebin.com/Cd3GqYbx
4) Copy the last file in application/helper folder as ckeditor_helper.php
5) Download the CKeditor controller here: http://pastebin.com/UD0bB9ig
6) Copy the controller in your application/controllers folder as ckeditor.php
7) Download the main ckeditor project from the official site: http://ckeditor.com/download/
8) Copy the ckeditor folder you just download into your asset folder (if you want you can also download the ckfinder project and put it in the same folder)
9) Add these line of js to your view file (adjust the path):

10) In your controller add this php code and adjust the path:
$this->load->library('ckeditor');
$this->load->library('ckfinder');

$this->ckeditor->basePath = base_url().'asset/ckeditor/';
$this->ckeditor->config['toolbar'] = array(
                array( 'Source', '-', 'Bold', 'Italic', 'Underline', '-','Cut','Copy','Paste','PasteText','PasteFromWord','-','Undo','Redo','-','NumberedList','BulletedList' )
                                                    );
$this->ckeditor->config['language'] = 'it';
$this->ckeditor->config['width'] = '730px';
$this->ckeditor->config['height'] = '300px';

//Add Ckfinder to Ckeditor
$this->ckfinder->SetupCKEditor($this->ckeditor,'../../asset/ckfinder/');

11) In your view print the editor with:
echo $this->ckeditor->editor("textarea name","default textarea value");

Comment: how to add image upload fucntionality? added Image in $config['toolbar'] but its not working. Please help
Comment: @user1149244 yes, you have to create it
Comment: @ Mansoorkhan Cherupuzha: I found this links helpful for me to make the image upload work in CKeditor http://derekmd.com/2013/03/ckeditor-image-upload-in-codeigniter/.  You might want to give it a try. One of the best tutorial I've found so far.
Comment: Works perfect using the listed steps!
Comment: I have follow your step, But I am facing issue that whenever I am click on Browse Server button that popup screen show me blank. Could you help me for that?
Comment: I have ask question here

http://stackoverflow.com/questions/20798373/browse-server-is-not-working-in-ckeditor-with-ckfinder
Comment: Everything works fine, except the CKEditor does not obey the height parameter. Why?
Comment: All your links are deleted ... :(
Comment: @ChristianGiupponi None of them works .... all are dead

see this http://pastebin.com/fkK9e0RR
Comment: Man, if I click on the link I see the code...All the links are set to EXPIRES: NEVER
Comment: @ChristianGiupponi :
Can You Help , how can I get value of ckeditor field in Controller . while it is having
'echo $this->ckeditor->editor("textarea name","default textarea value");'
to show ck editor , how can i submit value using this?
any idea ?
Comment: @DevangMistry you need to wrap it in a form and use the submit event to retrieve content in Controller
Comment: links not working. would you please correct the links?
Comment: is this method secure to insert in database?
Comment: @metalbox this will no insert any data into your database, it just show the steps to add CKEditor to textareas, the form submission is on your own
Answer: You could otherwise do this:

copy the CKEditor files to a folder in your source's root eg ckeditor
Include the CKEditor files in your view file

        <link rel="stylesheet" href="<?php base_url(); ?>style/format.css">

finally your textarea in your html document

 <textarea cols="80" id="edi" name="editor1" rows="10">
                <?php echo $page_content->message1; ?>
                            </textarea>

This works great for me.  Enjoy!
Answer: I found a sweetly simple 2-code-line explanation here:  http://www.iprogrammerindia.in/how-to-integrate-ckeditor-in-codeigniter/#comment-73
Just in case the link disappears I will paste the text here.  This worked for me 8/1/14:
Include this line in your view where you want to use ckeditor and place your ckeditor folder in root folder.  Here i placed in js/ckeditor/ in root folder

Next include the below line in the same view,
<?php echo form_textarea(array('name' =>'desc','id'=>'desc','class'=>"ckeditor")); ?>

Answer:
Download CKEditor package from https://ckeditor.com/ckeditor-4/download/
Unzip folder in your Codeigniter project folder at your preferred location.
Include the

        <form>
            <textarea name="editor1" id="editor1" rows="10" cols="80">
                This is my textarea to be replaced with CKEditor.
            </textarea>

        </form>

Comment: Thanks, more easy impossible
Answer: Well, I know this question is old, but this is what I did, and seems to be easiest for me.

In my root, I have a directory called 'js' and within that I have a
directory called 'plugins'. I copied the ckeditor files there.
Then in application/views/common/headers directory I have a header file
titled 'ckeditor.php'. Within this file just lies the following code:

Then in the controller, I added the header file to the $data object to pass to the view:
$data['header_files']  =   array('header1','header2','header3', 'ckeditor'); // header file names were changed for the sake of my client
Then, you pass the $data object of to the view of course:
$this->load->view('common/admin-template',$data);
then I just called CKEDITOR.replace('textareaNameHere');

and voila. it works
Answer: same problem is i am fessing but just little thing u need all file in asset folder in the root folder.
my controller part is
$this->load->library('ckeditor');
    $this->load->library('ckfinder');

    $this->ckeditor->basePath = base_url().'assets/admin/ckeditor/';
    $this->ckeditor->config['toolbar'] = array(
                                                array( 'Source', '-', 'Bold', 'Italic', 'Underline', '-','Cut','Copy','Paste','PasteText','PasteFromWord','-','Undo','Redo','-','NumberedList','BulletedList' )
                                                );
    $this->ckeditor->config['language'] = 'en';
    $this->ckeditor->config['width'] = '730px';
    $this->ckeditor->config['height'] = '300px';

//Add Ckfinder to Ckeditor
    $this->ckfinder->SetupCKEditor($this->ckeditor,base_url().'asset/admin/ckfinder/');

my view part is

    Description <?php echo form_error('event_description'); ?>
    <?php echo $this->ckeditor->editor("event_description",((isset($event_description)) ? $event_description : ''));?>

i putting ck editer folder in asset folder and asset folder in root file
like
C:\wamp\www\site\assets\admin\ckeditor
Comment: Is this an answer, or a new question?
Comment: it is answer of above question
Answer: I think the simplest way to use Ckeditor is through CDN, Use this in your view file and
<!DOCTYPE html>

<meta charset="utf-8">
<title>CKEditor</title>

<textarea name="editor" id="editor" rows="10" cols="80" placeholder="Insert text here" class="form-control"></textarea>

for more information and styling of ckeditor and use of different styles of ckeditor you can visit the https://cdn.ckeditor.com/#ckeditor5
If you are looking for Document editor of Ckeditor then follow the below link
https://ckeditor.com/docs/ckeditor5/latest/builds/guides/quick-start.html#document-editor
use the example and you will be able to insert the document editor in your view file

CKEditor in CodeIgniter