Q:

How to execute a script within bash script, and not wait for it to complete?

I am writing a script that runs occasionally and pulls jobs out of a mysql db.  When I do this, I run something like the following:
job="/tmp/blah.sh arg1 arg2 arg3"
eval $job

I need it to move on right after I run the eval, and not wait for the other script to complete though. What is an easy way to do that?  I tried
exec $job & 

based on a thread I found here, but that did the opposite, it ran the script, and then just stalled after the "job" completed, and stopped my entire script.
EDIT:
The problem I am running in to with separating my args and script is that jobs has multiple lines and looks like:
/tmp/blah1.sh arg1 arg2 arg3 arg4
/tmp/blah2.sh null null null arg4
/tmp/blah3.sh arg1 null arg3 arg4

So currently I have it running just : eval $jobs : if there is only one line, and if there are multiples, I do a for loop and run each line.  What is the best way to run this in a for loop to pull out and separate the args and scripts?

A:

I wouldn't use eval; instead, separate the command name from the arguments, then invoke the command in a normal fashion.
jobCmd="/tmp/blah.sh"
jobArguments=( "arg1" "arg2" "arg3" )

$jobCmd "${jobArguments[@]}" &

