how do I do the following from a shell script (run several commands at once and wait for them)?

Question

I want to write a shell scripts that executes a few commands and waits for all of them to terminate.

I think what I would have to do is use

      cmd1 &
      cmd2 &
      cmd3 &
      ....

etc.

but what I don't know is how to wait for them to terminate.

any ideas?

FatalError · Answer 1 · 2012-09-27T17:59:16.277

1

Just use the wait command without any args:

cmd1 &
cmd2 &
cmd3 &
wait

This will make bash wait for all unfinished children. This will work as long as you don't have other background tasks that you need to continue as well.

edited Sep 27 '12 at 17:59

answered Sep 27 '12 at 17:52

FatalError

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score 0 · Accepted Answer · edited May 23 '17 at 11:48

0

This question has the answer to your problem. :)

You would simply wait for all running jobs by calling the wait command.

How to wait in bash for several subprocesses to finish and return exit code !=0 when any subprocess ends with code !=0?

edited May 23 '17 at 11:48

Community

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answered Sep 27 '12 at 17:40

Gung Foo

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";" wouldn't work for me because I want them to all run together in the background. Kind of like forking and waiting for all the processes you forked to terminate. – kloop Sep 27 '12 at 17:43

score 0 · Answer 3 · answered Sep 27 '12 at 18:13

This example seems to work.

[stou$ ~]$ (sleep 1 ; echo 1) & (sleep 4 ; echo 4) & (sleep 3 ; echo 3)&  wait; echo done
[1] 13651
[2] 13652
[3] 13653
1
[1]   Done                    ( sleep 1; echo 1 )
3
4
[2]-  Done                    ( sleep 4; echo 4 )
[3]+  Done                    ( sleep 3; echo 3 )
done
[stou$ ~]$

how do I do the following from a shell script (run several commands at once and wait for them)?

3 Answers3