Bash: Call shell script but do not wait for return code

Question

Is it possible in Bash to call a shell script from another shell script but not have the original script wait for the sub-script to complete?

Nathan C · Accepted Answer · 2013-06-18 13:21:26Z

7

Just fork it with a &. As in, sh /path/to/script/script.sh &

This will print messages from the subscript, but you can replace the & with >/dev/null & and suppress the output.

answered Jun 18, 2013 at 13:21

Nathan C

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So the original script will continue on and can complete with a 0 exit code regardless of if the called script is still running, correct? What if set -e is invoked, does it make a difference?
– user160910
Jun 18, 2013 at 14:59
It doesn't. Once it's forked, it's on its own process and PID. nohup as mentioned by the other answer will also keep the process going through logouts and such.
– Nathan C
Jun 18, 2013 at 16:12
right after you fork the process, you can "capture" its PID with $!. E.g. mypid=$! . I recommend then using wait $mypid at some point in your script to ensure that the forked process ends before ending the script.
– senorsmile
Jun 19, 2013 at 3:29

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Pascal Schmiel · Accepted Answer · 2013-06-18 13:53:18Z

1

You should use "nohup" to make sure the process / script completes even if your user is logged out:

nohup /my/script.sh &

answered Jun 18, 2013 at 13:53

Pascal Schmiel

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