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

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```

其他回答

我已经尝试过了,并结合了其他例子中最好的部分。该脚本将在任何后台进程退出时执行checkpid函数,并输出退出状态而不诉诸轮询。

#!/bin/bash

set -o monitor

sleep 2 &
sleep 4 && exit 1 &
sleep 6 &

pids=`jobs -p`

checkpids() {
    for pid in $pids; do
        if kill -0 $pid 2>/dev/null; then
            echo $pid is still alive.
        elif wait $pid; then
            echo $pid exited with zero exit status.
        else
            echo $pid exited with non-zero exit status.
        fi
    done
    echo
}

trap checkpids CHLD

wait

为了将此并行化…

for i in $(whatever_list) ; do
   do_something $i
done

翻译成这样…

for i in $(whatever_list) ; do echo $i ; done | ## execute in parallel...
   (
   export -f do_something ## export functions (if needed)
   export PATH ## export any variables that are required
   xargs -I{} --max-procs 0 bash -c ' ## process in batches...
      {
      echo "processing {}" ## optional
      do_something {}
      }' 
   )

If an error occurs in one process, it won't interrupt the other processes, but it will result in a non-zero exit code from the sequence as a whole. Exporting functions and variables may or may not be necessary, in any particular case. You can set --max-procs based on how much parallelism you want (0 means "all at once"). GNU Parallel offers some additional features when used in place of xargs -- but it isn't always installed by default. The for loop isn't strictly necessary in this example since echo $i is basically just regenerating the output of $(whatever_list). I just think the use of the for keyword makes it a little easier to see what is going on. Bash string handling can be confusing -- I have found that using single quotes works best for wrapping non-trivial scripts. You can easily interrupt the entire operation (using ^C or similar), unlike the the more direct approach to Bash parallelism.

下面是一个简化的工作示例……

for i in {0..5} ; do echo $i ; done |xargs -I{} --max-procs 2 bash -c '
   {
   echo sleep {}
   sleep 2s
   }'

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

陷阱是你的朋友。在很多系统中都可能出现ERR。您可以捕获EXIT,或在DEBUG上在每个命令之后执行一段代码。

这除了所有的标准信号。

edit

这是一次意外的登录错误的帐户,所以我没有看到示例的请求。

试试这里,用我的普通账户。

在bash脚本中处理异常

我需要这个,但目标进程不是当前shell的子进程,在这种情况下,等待$PID不起作用。我确实找到了以下替代方案:

while [ -e /proc/$PID ]; do sleep 0.1 ; done

这依赖于procfs的存在,它可能不可用(例如Mac不提供它)。所以对于可移植性,你可以用这个代替:

while ps -p $PID >/dev/null ; do sleep 0.1 ; done