How to trap on exit 1 only in bash

Question

I have the following bash script:

#!/bin/bash


function hello {
    echo "Hello World!"
}

trap hello EXIT

The problem is that this function will execute on any exit code for that script. if I append an exit 1 at the end of the script, Hello World! will be printed, but it also will get printed on exit 0.

Is there a way to run a function within a script, but only if the script exits with exit code 1 only? (actually anything other than 0 works too.)

Here is a hint, replace `Hello World!` with `$?` and see what it prints. — oguz ismail, Dec 23 '20 at 07:58

score 5 · Answer 1 · answered Dec 23 '20 at 14:27

5

Capture the current status in the function:

bye() {
  local status=$?
  (( status == 1 )) && echo "Hello World!"
  echo "Exiting with status $status"
  exit $status
}
trap bye EXIT
# ...
exit $some_status

answered Dec 23 '20 at 14:27

glenn jackman

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score 3 · Accepted Answer · edited Dec 23 '20 at 11:29

3

This will work, but with losing the exit code in the function:

trap '[[ $? -eq 1 ]] && hello' EXIT

If you need the exit code in the function, you have to do the test inside the function.

edited Dec 23 '20 at 11:29

oguz ismail

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answered Dec 23 '20 at 11:22

Marco

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How to trap on exit 1 only in bash

2 Answers2