3

My bash script (init.sh) call another script (script.sh) and I want to test the error code from script.sh before doing any further action in init.sh.
I thought about testing it with $?, but it does not work

My init.sh is like the following:

#!/bin/bash
set -e
echo "Before call"
docker run -v $PWD:/t -w /t [command]
if [ $? == 1 ]; then
        echo "Issue"
fi
echo "After call"

I only got the Before call from stdout and not the After call.

I know for a fact that if I execute docker run -v $PWD:/t -w /t [command] alone with wrong arguments, then echo $? will rightly display 1.

I was thinking that I do not catch the exit code from scrip.sh, but from somewhere else.

Any ideas?

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2
  • 1
    This is confusing ... you mention script.sh but don't call it in your init.sh. Anyhow, you might try run bash -x init.sh and see what it's actually doing. Commented Sep 29, 2016 at 14:59
  • 1
    With set -e, if the call to docker does set $? to 1, your script will exit before the if statement executes, let alone the final echo call. Commented Sep 29, 2016 at 15:01

3 Answers 3

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5

You running the script with set -e. This means that if any command exits with a non zero status, bash will stop executing all subsequent lines. So here, if docker exits with status 1, the conditional that follows will not have a chance to run at all. Try this instead:

#!/bin/bash
set -e
echo "Before call"
if ! docker run -v $PWD:/t -w /t [command]; then
        echo "Issue"
fi
echo "After call"

This runs the command inside the if test which suppresses the effect of set -e I described above and gives you a chance to catch the error. Note this is will also catch all non-zero statuses, not just 1.

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6 Comments

set -e is not generally a good idea.
@chepner You are right. In fact, I guess this question itself is an example why.
The question just seems to overlook that set -e was in effect. The actual problems with set -e involve more unexpected exits (or failures to exit).
So, what's the more efficient way? Get rid of the set -e or use the solution from @redneb?
Just get rid of set -e. If you really need the script to exit, you can call exit explicitly (with the appropriate exit status) from within the if statement.
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1

Bash numeric comparison operator is -eq, and not ==...
So:

#!/bin/bash
set -e
echo "Before call"
docker run -v $PWD:/t -w /t [command]
if [ $? -eq 1 ]; then
        echo "Issue"
fi
echo "After call"
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4 Comments

It's somewhat irrelevant; there aren't any values that $?could have that would cause $? -eq 1 and $? = 1 (use = with [, not ==) to behave differently.
@Mornor: why did it fail? Because the command did not exit with -1? Or why? You should echo the value of $?, to test it's value...
Because I set -e, you can check the answer from @redneb
I see. However, the syntax was incorrect in the first place. The solution for you is probably not to use -e flag... :-)
1

set -e is generally a bad idea. Sure, it may seem like a good idea to have your script exit automatically in the event of an unexpected error, but the problem is that set -e and you may have different ideas about what constitutes a fatal error.

Instead, do your own error handling.

#!/bin/bash
echo "Before call"
docker run -v $PWD:/t -w /t [command]
docker_status=$?
if [ $docker_status != 0 ]; then
    echo "docker returned: $docker_status"
    exit $docker_status
fi
echo "After call"

In this simple code, I've somewhat redundantly saved the value of $? to another variable first. This ensures that it is preserved after you start executing other commands that examine, log, or otherwise process the value of $?. Also, I'm logging and exiting here on any non-zero status, not just 1. In theory, you might take different action for an exit status of 1 than for an exit status of 2, but here we take the same log-then-exit action for any error.

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