How to trap exit code in Bash script
The accepted answer is basically correct, I just want to clarify things.
The following example works well:
#!/bin/bashcleanup() { rv=$? rm -rf "$tmpdir" exit $rv}tmpdir="$(mktemp)"trap "cleanup" EXIT# Do things...
But you have to be more careful if doing cleanup inline, without a function. For example this won't work:
trap "rv=$?; rm -rf $tmpdir; exit $rv" EXIT
Instead you have to escape the $rv
and $?
variables:
trap "rv=\$?; rm -rf $tmpdir; exit \$rv" EXIT
You might also want to escape $tmpdir
, as it will get evaluated when the trap line gets executed and if the tmpdir
value changes later that might not give the expected behaviour.
Edit: Use shellcheck to check your bash scripts and be aware of problems like this.
I've found it is better to separate EXIT trap from the trap for other signals
Example trap test script...
umask 77tmpfile=`tmpfile.$$`trap 'rm -f "$tmpfile"' EXITtrap 'exit 2' HUP INT QUIT TERMtouch $tmpfileread -r input exit 10
The temporary file is cleaned up.The file exit value of 10 is preserved!Interrupts result in an exit value of 2
Basically as long as you don't use "exit" in a EXIT trap, it will exit with the original exit value preserved.
ASIDE: Note the quoting in the EXIT trap. That lets me change what file needs to be cleaned up during the scripts lifetime. I often also include a test for the existence of the $tmpfile before trying to remove it, so I don't even need to set it at the start of the script, only before creating it.