I'm trying to get the output of env in a shell variable and print it.#!/bin/shENV_BEFORE=$(env)printf $ENV_BEFOREAs a result a single variable from the env output gets printed.When using echo instead of printf all the output is printed, however without the newlines.What am I missing here?
Printf printing only a single line
bash;shell;quoting;printf
The problem is that you're not quoting the $ENV variable. As explained in man bash:Enclosing characters in double quotes preserves the  literal  value  of all  characters  within the quotes, with the exception of $, `,  \, and,    when history expansion is enabled, !.  The characters $   and  `  retain    their  special meaning within double quotes.  The  backslash retains its    special meaning only when followed by one of  the following  characters:    $,  `,  , \, or .So, enclosing a sequence like \n in double quotes preserves its meaning. This is why, when not quoted, \n is just a normal n:$ printf \nn$While, when quoted:$ printf \n$An unquoted variable in bash invokes the split+glob operator. This means that the variable is split on whitespace (or whatever the special variable $IFS has been set to) and each resulting word is used as a glob (it will expand to match any matching file names). Your problem is with the split part of this. To illustrate, let's take a simpler multiline variable:$ var=$(printf foo\nbar\n)Now, using the shell's set -x debug feature, you can see exactly what's going on:$ echo $var+ echo foo barfoo bar$ echo $var+ echo 'foobar'foobarAs you can see above, echo $var (unquoted) subjects $var to split+glob so it results in two separate strings, foo and bar. The newline was eaten by th split+glob. When the variable was quoted, it wasn't subjected to split+glob, the newline was kept and, because it is quoted, is also interpreted correctly ad printed out.The next problem is that printf is not like echo. It doesn't just print anything you give it, it expects a format string. For example printf color:%s green will print color:green because the %s will be replaced with green. It also ignores any input that can't fit into the format string it was given. So, if you run printf foo bar, printf will treat foo as its format string and bar as the variable it is supposed to format with it. Since there is no %s or equivalent to be replaced by bar, bar is ignored and foo alone is printed:$ printf $var+ printf foo barfooThat's what happened when you ran printf $ENV_BEFORE. Because the variable wasn't quoted, the split glob effectively replaced newlines with spaces, and printf only printed the first word it saw. To do it correctly, use format strings, and always quote your variables:printf '%s\n' $ENV_BEFORE