Why is exit my status valid in command line but not within bash script? (Bash)

Question

There are a few layers here, so bear with me.

My docker-container ssh -c"echo 'YAY!'; exit 25;" command executes echo 'YAY!'; exit 25; in my docker container. It returns:

YAY
error:  message=YAY!
, code=25

I need to know if the command within the container was successful, so I append the following to the command:

docker-container ssh -c"echo 'YAY!'; exit 25;"  >&1 2>/tmp/stderr; cat /tmp/stderr | grep 'code=' | cut -d'=' -f2 | { read exitStatus; echo $exitStatus; }

This sends the stderr to /tmp/stderr and, with the echo $exitStatus returns:

YAY!
25

So, this is exactly what I want. I want the $exitStatus saved to a variable. My problem is, I am placing this into a bash script (GIT pre-commit) and when this exact code is executed, the exit status is null.

Here is my bash script:

# .git/hooks/pre-commit

if [ -z ${DOCKER_MOUNT+x} ];
then
    docker-container ssh -c"echo 'YAY!'; exit 25;"  >&1 2>/tmp/stderr; cat /tmp/stderr | grep 'code=' | cut -d'=' -f2 | { read exitStatus; echo $exitStatus; }

    exit $exitStatus;
else
    echo "Container detected!"
fi;

Eh? You can't possibly be getting null as an exit status; the only possible values are integers. You might have 0, but you can't have null. — Charles Duffy
– Charles Duffy, Commented Feb 16, 2018 at 20:32
This is in a respect duplicative of bash: why piping to read only works when fed into while read construct, insofar as understanding the answer to the other explains why this doesn't work. BashFAQ #24 is pertinent reading as well. — Charles Duffy
– Charles Duffy, Commented Feb 16, 2018 at 20:37
BTW, writing to /tmp/stderr is dangerous if you're on a shared system. Two scripts using that same name at once is a problem, but a much worse problem is if another user (or a compromised network service -- and remember, /tmp is world-writable!) replaces /tmp/stderr with a symlink to a file you have permission to write to but they don't, and which they want to see deleted. Don't use hardcoded temporary filenames -- this is what mktemp exists to avoid. — Charles Duffy
– Charles Duffy, Commented Feb 16, 2018 at 20:40
BTW, cat foo | grep bar is significantly less efficient than grep bar <foo -- and cat foo | grep bar | cut -d= -f2 can be replaced with, say, awk -F= '/bar/ { print $2 }' <foo to do everything in one process. — Charles Duffy
– Charles Duffy, Commented Feb 16, 2018 at 20:44

glenn jackman · Accepted Answer · 2018-02-16 21:33:22Z

1

That's because you're setting the variable in a pipeline. Each command in the pipeline is run in a subshell, and when the subshell exits the variable are no longer available.

bash allows you to run the pipeline's last command in the current shell, but you also have to turn off job control

An example

# default bash
$ echo foo | { read x; echo x=$x; } ; echo x=$x
x=foo
x=

# with "lastpipe" configuration
$ set +m; shopt -s lastpipe

$ echo foo | { read x; echo x=$x; } ; echo x=$x
x=foo
x=foo

Add set +m; shopt -s lastpipe to your script and you should be good.

And as Charles comments, there are more efficient ways to do it. Like this:

source <(docker-container ssh -c "echo 'YAY!'; exit 25;" 2>&1 1>/dev/null | awk -F= '/code=/ {print "exitStatus=" $2}')
echo $exitStatus

edited Feb 16, 2018 at 21:33

answered Feb 16, 2018 at 21:25

glenn jackman

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