In a Bash script, how can I exit the entire script if a certain condition occurs?

user_39036 0
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I'm writing a script in Bash to test some code. However, it seems silly to run the tests if compiling the code fails in the first place, in which case I'll just abort the tests.

Is there a way I can do this without wrapping the entire script inside of a while loop and using breaks? Something like a dun dun dun goto?

exit-code exit scripting bash
8 Answers

I have the same question but cannot ask it because it would be a duplicate.

The accepted answer, using exit, does not work when the script is a bit more complicated. If you use a background process to check for the condition, exit only exits that process, as it runs in a sub-shell. To kill the script, you have to explicitly kill it (at least that is the only way I know).

Here is a little script on how to do it:

#!/bin/bash

boom() {
    while true; do sleep 1.2; echo boom; done
}

f() {
    echo Hello
    N=0
    while
        ((N++ <10))
    do
        sleep 1
        echo $N
        #        ((N > 5)) && exit 4 # does not work
        ((N > 5)) && { kill -9 $$; exit 5; } # works 
    done
}

boom &
f &

while true; do sleep 0.5; echo beep; done

This is a better answer but still incomplete a I really don't know how to get rid of the boom part.

user_41429840

You can close your program by program name on follow way:

for soft exit do

pkill -9 -x programname # Replace "programmname" by your programme

for hard exit do

pkill -15 -x programname # Replace "programmname" by your programme

If you like to know how to evaluate condition for closing a program, you need to customize your question.

user_152034610
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