PROBLEM:
I have a shell program that I have been writing but I can’t find out how to make sure that trap is trapping for cleanup at the end or because of a error in some command, it cleans up either way.
Here is the code:
################################### Successful exit then this cleanup ###########################################################3
successfulExit()
{
IFS=$IFS_OLD
cd "$HOME" || { echo "cd $HOME failed"; exit 155; }
rm -rf /tmp/svaka || { echo "Failed to remove the install directory!!!!!!!!"; exit 155; }
}
###############################################################################################################################33
####### Catch the program on successful exit and cleanup
trap successfulExit EXIT
QUESTION:
How can I make trap only trap EXIT on program finish?
Here is the full script:
Answer
On entry to the EXIT trap, $? contains the exit status. That’s the same value you’d find as $? after calling this script in another shell: either the argument passed to exit (truncated to the range 0–255) or the return status of the preceding command. In the case of an exit due to set -e, it’s the return status of the command that triggered the implicit exit.
Usually you should save $? and exit again with the same status.
cleanup () {
if [ -n "$1" ]; then
echo "Aborted by $1"
elif [ $status -ne 0 ]; then
echo "Failure (status $status)"
else
echo "Success"
fi
}
trap 'status=$?; cleanup; exit $status' EXIT
trap 'trap - HUP; cleanup SIGHUP; kill -HUP $$' HUP
trap 'trap - INT; cleanup SIGINT; kill -INT $$' INT
trap 'trap - TERM; cleanup SIGTERM; kill -TERM $$' TERM