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?

score 6 · Accepted Answer · answered Jun 18 '13 at 13:21

6

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 '13 at 13:21

Nathan C

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? – Jun 18 '13 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 '13 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 '13 at 03:29

score 1 · Answer 2 · answered Jun 18 '13 at 13:53

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 '13 at 13:53

Pascal Schmiel

2 Answers2