如何在bash脚本中等待从该脚本派生的几个子进程完成，然后在任何子进程以code !=0结束时返回退出代码!=0?
简单的脚本:
#!/bin/bash
for i in `seq 0 9`; do
doCalculations $i &
done
wait
上面的脚本将等待所有10个子进程，但它总是给出退出状态0(参见help wait)。我如何修改这个脚本，以便它将发现衍生子进程的退出状态，并在任何子进程以code !=0结束时返回退出代码1 ?
有没有比收集子进程的pid、按顺序等待它们并求和退出状态更好的解决方案呢?
当前回答
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
其他回答
下面的代码将等待所有计算的完成，并在任何doccalculation失败时返回退出状态1。
#!/bin/bash
for i in $(seq 0 9); do
(doCalculations $i >&2 & wait %1; echo $?) &
done | grep -qv 0 && exit 1
这是我目前想到的。我想了解如何在子进程终止时中断sleep命令，这样就不必根据使用情况调优WAITALL_DELAY。
waitall() { # PID...
## Wait for children to exit and indicate whether all exited with 0 status.
local errors=0
while :; do
debug "Processes remaining: $*"
for pid in "$@"; do
shift
if kill -0 "$pid" 2>/dev/null; then
debug "$pid is still alive."
set -- "$@" "$pid"
elif wait "$pid"; then
debug "$pid exited with zero exit status."
else
debug "$pid exited with non-zero exit status."
((++errors))
fi
done
(("$#" > 0)) || break
# TODO: how to interrupt this sleep when a child terminates?
sleep ${WAITALL_DELAY:-1}
done
((errors == 0))
}
debug() { echo "DEBUG: $*" >&2; }
pids=""
for t in 3 5 4; do
sleep "$t" &
pids="$pids $!"
done
waitall $pids
只需将结果存储在shell之外，例如在一个文件中。
#!/bin/bash
tmp=/tmp/results
: > $tmp #clean the file
for i in `seq 0 9`; do
(doCalculations $i; echo $i:$?>>$tmp)&
done #iterate
wait #wait until all ready
sort $tmp | grep -v ':0' #... handle as required
等待所有作业并返回最后一个失败作业的退出码。与上面的解决方案不同，这不需要保存pid，也不需要修改脚本的内部循环。走开，等着吧。
function wait_ex {
# this waits for all jobs and returns the exit code of the last failing job
ecode=0
while true; do
[ -z "$(jobs)" ] && break
wait -n
err="$?"
[ "$err" != "0" ] && ecode="$err"
done
return $ecode
}
编辑:修正了脚本运行不存在的命令时可能被愚弄的错误。
这是我使用的东西:
#wait for jobs
for job in `jobs -p`; do wait ${job}; done
