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 ?

Simple script:

#!/bin/bash
for i in `seq 0 9`; do
  doCalculations $i &
done
wait

The above script will wait for all 10 spawned subprocesses, but it will always give exit status 0 (see help wait). How can I modify this script so it will discover exit statuses of spawned subprocesses and return exit code 1 when any of subprocesses ends with code !=0?

Is there any better solution for that than collecting PIDs of the subprocesses, wait for them in order and sum exit statuses?

1
494
5/2/2018 9:47:34 PM

Accepted Answer

wait also (optionally) takes the PID of the process to wait for, and with $! you get the PID of the last command launched in background. Modify the loop to store the PID of each spawned sub-process into an array, and then loop again waiting on each PID.

# run processes and store pids in array
for i in $n_procs; do
    ./procs[${i}] &
    pids[${i}]=$!
done

# wait for all pids
for pid in ${pids[*]}; do
    wait $pid
done
448
6/25/2018 10:36:50 AM

http://jeremy.zawodny.com/blog/archives/010717.html :

#!/bin/bash

FAIL=0

echo "starting"

./sleeper 2 0 &
./sleeper 2 1 &
./sleeper 3 0 &
./sleeper 2 0 &

for job in `jobs -p`
do
echo $job
    wait $job || let "FAIL+=1"
done

echo $FAIL

if [ "$FAIL" == "0" ];
then
echo "YAY!"
else
echo "FAIL! ($FAIL)"
fi

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