Bash ignore error and get return code

Question

I am using set -e to abort on errors.

But for particular one function I want to ignore error and on error I want return code of the function.

Example:

do_work || true 
 if [ $? -ne 0 ] 
  then
   echo "Error"
  fi  

But it is not work return code is always true due || true

How to get return code on do_work on error ?

Answer 1

You could use a subshell shortcut:

( set +e; do_work )
 if [ $? -ne 0 ]
  then
   echo "Error"
  fi

Hope this helps =)

Answer 2

Several of the answers given here are not correct, because they result in a test against a variable that will be un-defined if do_work succeeds.

We need to cover the successful case as well, so the answer is:

set -eu
do_work && status=0 || status=1

The poster's question is a little ambiguous because it says in the text "on error I want return code" but then the code implies "I always want the return code"

To illustrate, here is problematic code:

set -e

do_work() {
    return 0
}

status=123

do_work || status=$?
echo $status

In this code the value printed is 123, and not 0 as we might hope for.