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I am trying to create a script which will start many background command. For each background command I need to get the return code.

I have been trying the following script :

set -x

for i in 1 2
 echo start $i
 ssh mysql "/root/test$i.sh" &

for i in ${#pid[@]}
echo ${pid[$i]}
wait ${pid[$i]}

if [ ${return[$i]} -ne 0 ]
  echo mail error


echo ${return[1]}
echo ${return[2]}

My issue is during the wait loop, if the second pid finish before the first one, I'll not be able to get the return code.

I know that I can run wait pid1 pid2, but with this command I can't get the return code of all commands.

Any idea ?

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2 Answers 2

up vote 1 down vote accepted

You can do this by using a temporary directory.

# Create a temporary directory to store the statuses
dir=$(mktemp -d)

# Execute the backgrouded code. Create a file that contains the exit status.
# The filename is the PID of this group's subshell.
for i in 1 2; do
    { ssh mysql "/root/test$i.sh" ; echo "$?" > "$dir/$BASHPID" ; } &

# Wait for all jobs to complete

# Get return information for each pid
for file in "$dir"/*; do
    printf 'PID %d returned %d\n' "${file##*/}" "$(<"$file")"

# Remove the temporary directory
rm -r "$dir"
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The issue is more with your

for i in ${#pid[@]}

Which is for i in 2.

It should rather be:

for i in 1 2


for ((i = 1; i <= ${#pid[@]}; i++))

wait "$pid" will return the exit code of the job with bash (and POSIX shells, but not zsh) even if the job had already terminated when wait was started.

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