multiple command prioritization in bash

Question

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

I have the following problem:

I have 3 processes that I put in a script

process1 process2 process3

I want that process 1 and 2 run simultaneously but they have both finished before the process 3 starts.

i guess it is something like the following..but I am not sure about that "wait"

#!/bin/sh

    (
      process1 &
      process2 &

      wait

      process3

                )

Thank you

Fabio

That will do what you want. You do not need the ( ) construct. — jim mcnamara, Feb 05 '13 at 11:13

Davide Berra · Accepted Answer · 2013-02-05T11:33:01.497

3

Just save the pid of both processes and wait for both to exit

#!/bin/bash

process1 &
pid1=$!
process2 &
pid2=$!

wait ${pid1}
echo "Return value of process1: $?"
wait ${pid2}
echo "Return value of process2: $?"

process3

edited Feb 05 '13 at 11:33

answered Feb 05 '13 at 11:21

Davide Berra

6,387
2
29
50

so the single wait does not make any sense in your opinion? – fabioln79 Feb 05 '13 at 11:25
It works as well. This is just my habit and I think it's better because, if you want, you can check the return value of the processes – Davide Berra Feb 05 '13 at 11:29

multiple command prioritization in bash

1 Answers1