Skip 'natural' exit in trap

Question

My script looks like:

trap "$(pwd)/dd-destroy.sh $l-env;$(pwd)/dd-drop.sh $l-db;exit" INT QUIT TERM EXIT

./dd-all.sh $l $l-db $l-env || exit 1

app_ip=$(./dd-status.sh ip $l-env|grep docker-app|awk '{print $3}')
url=http://$app_ip:8080/app/

wget -O /dev/null $url || (echo "access $url failed" && false) || exit 1

Now I want this trap launch when error occurs, those 'exit 1' (I can change them there). But when nothing happens, like dd-all.sh worked fine, when wget came fine. I do not want to call for destroy.

Can I accomplish this with trap? It's not my script, I need to change it a little. Maybe trap is a bad call here?

Answer 1

Simply change the start of your trap to check for the return code of the last command (which will be that of the exit, or the command at the end of the script) for zero. I.e.

trap '[ $? = 0 ] && exit;'"..."

where "..." is your current string, which should follow with no intervening spaces. An exit within the trap will not trap again.

If your trap is also handling signals, eg SIGINT (control-C), then the trap is called once for the SIGINT, then once more for the exit. You can probably check if trap is being called from SIGINT as the $? will be 128+n where n is the signal, so 130 for SIGINT. Remember $? will be changed as soon as you give another command, so perhaps start with rc=$?; then use $rc in your tests.