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。

其他回答

如果你安装了GNU Parallel,你可以这样做:

# If doCalculations is a function
export -f doCalculations
seq 0 9 | parallel doCalculations {}

GNU Parallel会给你退出代码:

0 -所有作业运行无错误。 1-253 -部分作业失败。退出状态给出了失败作业的数量 254—超过253个作业失败。 255 -其他错误。

观看介绍视频了解更多信息:http://pi.dk/1

wait还(可选地)接受要等待的进程的PID,并且使用$!你会得到最后一个命令的PID在后台启动。 修改循环,将每个衍生子进程的PID存储到一个数组中,然后再次循环等待每个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

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
   }'

我几乎陷入了使用jobs -p来收集pid的陷阱,如果子进程已经退出,这将不起作用,如下面的脚本所示。我选择的解决方案是简单地调用-n N次,其中N是我有孩子的数量,这是我确定知道的。

#!/usr/bin/env bash

sleeper() {
    echo "Sleeper $1"
    sleep $2
    echo "Exiting $1"
    return $3
}

start_sleepers() {
    sleeper 1 1 0 &
    sleeper 2 2 $1 &
    sleeper 3 5 0 &
    sleeper 4 6 0 &
    sleep 4
}

echo "Using jobs"
start_sleepers 1

pids=( $(jobs -p) )

echo "PIDS: ${pids[*]}"

for pid in "${pids[@]}"; do
    wait "$pid"
    echo "Exit code $?"
done

echo "Clearing other children"
wait -n; echo "Exit code $?"
wait -n; echo "Exit code $?"

echo "Waiting for N processes"
start_sleepers 2

for ignored in $(seq 1 4); do
    wait -n
    echo "Exit code $?"
done

输出:

Using jobs
Sleeper 1
Sleeper 2
Sleeper 3
Sleeper 4
Exiting 1
Exiting 2
PIDS: 56496 56497
Exiting 3
Exit code 0
Exiting 4
Exit code 0
Clearing other children
Exit code 0
Exit code 1
Waiting for N processes
Sleeper 1
Sleeper 2
Sleeper 3
Sleeper 4
Exiting 1
Exiting 2
Exit code 0
Exit code 2
Exiting 3
Exit code 0
Exiting 4
Exit code 0

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。