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
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 $pids
其他回答
如果您有bash 4.2或更高版本可用，以下内容可能对您有用。它使用关联数组存储任务名称及其“代码”，以及任务名称及其pid。我还内置了一个简单的速率限制方法，如果你的任务消耗了大量CPU或I/O时间，你想限制并发任务的数量，这个方法可能会派上用场。
脚本在第一个循环中启动所有任务，在第二个循环中使用结果。
对于简单的情况，这有点过分，但它允许非常简洁的东西。例如，可以将每个任务的错误消息存储在另一个关联数组中，并在一切都解决后打印它们。
#! /bin/bash
main () {
local -A pids=()
local -A tasks=([task1]="echo 1"
[task2]="echo 2"
[task3]="echo 3"
[task4]="false"
[task5]="echo 5"
[task6]="false")
local max_concurrent_tasks=2
for key in "${!tasks[@]}"; do
while [ $(jobs 2>&1 | grep -c Running) -ge "$max_concurrent_tasks" ]; do
sleep 1 # gnu sleep allows floating point here...
done
${tasks[$key]} &
pids+=(["$key"]="$!")
done
errors=0
for key in "${!tasks[@]}"; do
pid=${pids[$key]}
local cur_ret=0
if [ -z "$pid" ]; then
echo "No Job ID known for the $key process" # should never happen
cur_ret=1
else
wait $pid
cur_ret=$?
fi
if [ "$cur_ret" -ne 0 ]; then
errors=$(($errors + 1))
echo "$key (${tasks[$key]}) failed."
fi
done
return $errors
}
main
我真的很喜欢卢卡的回答，但需要它为zsh，所以这里是作为参考:
pids=()
# run processes and store pids in array
for i in $n_procs; do
./procs[${i}] &
pids+=($!)
done
# wait for all pids
for pid in ${pids[*]}; do
wait $pid
done```
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 $pids
我最近用过这个(感谢Alnitak):
#!/bin/bash
# activate child monitoring
set -o monitor
# locking subprocess
(while true; do sleep 0.001; done) &
pid=$!
# count, and kill when all done
c=0
function kill_on_count() {
# you could kill on whatever criterion you wish for
# I just counted to simulate bash's wait with no args
[ $c -eq 9 ] && kill $pid
c=$((c+1))
echo -n '.' # async feedback (but you don't know which one)
}
trap "kill_on_count" CHLD
function save_status() {
local i=$1;
local rc=$2;
# do whatever, and here you know which one stopped
# but remember, you're called from a subshell
# so vars have their values at fork time
}
# care must be taken not to spawn more than one child per loop
# e.g don't use `seq 0 9` here!
for i in {0..9}; do
(doCalculations $i; save_status $i $?) &
done
# wait for locking subprocess to be killed
wait $pid
echo
从这里，我们可以很容易地推断，并拥有一个触发器(触摸文件，发送信号)并改变计数标准(计数触摸的文件，或其他)以响应该触发器。或者如果你只是想要'any'非零rc，只需从save_status中杀死锁。
