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

do_work || {
    status=$?
    echo "Error"
}

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.

Answer 3

You could use a subshell shortcut:

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

Hope this helps =)

Answer 4

One way is to use a pipe, -e only looks at the right-most result of a pipe:

set -e

do_work | true

retn=${PIPESTATUS[0]}
if (( $retn != 0 ))
then   
    echo "Error $retn"
fi     
echo Ending

I wrote a simple do_work which just did exit 42 and got the following output:

Error 42
Ending

The PIPESTATUS array is maintained by Bash, with each element giving the return code of each part of the pipeline. We need to capture it at once (hence $retn) since it is overwritten at each command.

Of course this might be problematic if your do_work includes a pipe itself.

Answer 5

do_work || status=$?
if [ $status -ne 0 ]
then
    echo "Oh no - Fail whale $status has arrived"
fi