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

Question

How to wait in a bash script for several subprocesses spawned from that script to finish and return exit code !=0 when any of the subprocesses ends with code !=0 ?

S

Answer 1

I think that the most straight forward way to run jobs in parallel and check for status is using temporary files. There are already a couple similar answers (e.g. Nietzche-jou and mug896).

#!/bin/bash
rm -f fail
for i in `seq 0 9`; do
  doCalculations $i || touch fail &
done
wait 
! [ -f fail ]

The above code is not thread safe. If you are concerned that the code above will be running at the same time as itself, it's better to use a more unique file name, like fail.$$. The last line is to fulfill the requirement: "return exit code 1 when any of subprocesses ends with code !=0?" I threw an extra requirement in there to clean up. It may have been clearer to write it like this:

#!/bin/bash
trap 'rm -f fail.$$' EXIT
for i in `seq 0 9`; do
  doCalculations $i || touch fail.$$ &
done
wait 
! [ -f fail.$$ ] 

Here is a similar snippet for gathering results from multiple jobs: I create a temporary directory, story the outputs of all the sub tasks in a separate file, and then dump them for review. This doesn't really match the question - I'm throwing it in as a bonus:

#!/bin/bash
trap 'rm -fr $WORK' EXIT

WORK=/tmp/$$.work
mkdir -p $WORK
cd $WORK

for i in `seq 0 9`; do
  doCalculations $i >$i.result &
done
wait 
grep $ *  # display the results with filenames and contents