在等待流程之前，流程可能已经完成。如果我们触发等待一个已经完成的进程，它将触发一个错误，比如pid不是这个shell的子进程。为了避免这种情况，可以使用以下函数来查找过程是否完成:

isProcessComplete(){
PID=$1
while [ -e /proc/$PID ]
do
    echo "Process: $PID is still running"
    sleep 5
done
echo "Process $PID has finished"
}

这是我使用的东西:

#wait for jobs
for job in `jobs -p`; do wait ${job}; done

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。

使用'wait -n'来等待多个子进程，并在其中任何一个进程以非零状态码退出时退出。

#!/bin/bash
wait_for_pids()
{
    for (( i = 1; i <= $#; i++ )) do
        wait -n $@
        status=$?
        echo "received status: "$status
        if [ $status -ne 0 ] && [ $status -ne 127 ]; then
            exit 1
        fi
    done
}

sleep_for_10()
{
    sleep 10
    exit 10
}

sleep_for_20()
{
    sleep 20
}

sleep_for_10 &
pid1=$!

sleep_for_20 &
pid2=$!

wait_for_pids $pid2 $pid1

状态代码'127'是不存在的进程，这意味着子进程可能已经退出。

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。

从Bash 5.1开始，由于引入了wait -p，有了一种很好的等待和处理多个后台作业结果的新方法:

#!/usr/bin/env bash

# Spawn background jobs
for ((i=0; i < 10; i++)); do
    secs=$((RANDOM % 10)); code=$((RANDOM % 256))
    (sleep ${secs}; exit ${code}) &
    echo "Started background job (pid: $!, sleep: ${secs}, code: ${code})"
done

# Wait for background jobs, print individual results, determine overall result
result=0
while true; do
    wait -n -p pid; code=$?
    [[ -z "${pid}" ]] && break
    echo "Background job ${pid} finished with code ${code}"
    (( ${code} != 0 )) && result=1
done

# Return overall result
exit ${result}