Check output of bash script

Question

I am using the bash script wait for it to check if a container is up and running before I launch another container.

https://github.com/vishnubob/wait-for-it

It seems that if it's successful it returns 0 and if not 124. Is it possible to check this value and if it's not successful exit the script?

I've tried

./wait-for-it.sh $BROKER_ADDRESS echo $? if ($?==124) then echo "exiting as broker service never became available" exit fi

And I can see it echos 124 in my terminal but then the check fails and it carries on launching the container. I'm assuming my conditional check is wrong but I can't seem to figure out why

You need to save $? right after the call to wait-for-it.sh in a variable and use it. Otherwise, $? could be different - it represents the last command's exit status. — P.P
– P.P, Commented May 4, 2020 at 15:25
After the echo, $? is 0 because the echo succeeded. Use status=$? and then check $status, etc. Or test $? without echoing it, but capturing it gives you more flexibility. — Jonathan Leffler
– Jonathan Leffler, Commented May 4, 2020 at 15:25
An arithmetic conditional requires double parens, like (($? == 124)), by the way. Otherwise, you're trying to run $?==124 as a command in a subshell. — Benjamin W.
– Benjamin W., Commented May 4, 2020 at 15:29

user2700022 · Accepted Answer · 2020-05-04 18:27:39Z

$? Always returns the exit status of the previous command so in this case in your 3rd line in if loop you are comparing exit status of the echo command in your second line to 124. So either remove the second line or store the exit status to a variable & use that variable in your if loop.

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