这是我使用的东西:

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

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

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

我有一个类似的情况，但有各种各样的问题与循环子shell，确保这里的其他解决方案不能工作，所以我让我的循环编写脚本，我将运行，等待结束。有效:

#!/bin/bash
echo > tmpscript.sh
for i in `seq 0 9`; do
    echo "doCalculations $i &" >> tmpscript.sh
done
echo "wait" >> tmpscript.sh
chmod u+x tmpscript.sh
./tmpscript.sh

愚蠢，但简单，并帮助调试一些事后的事情。

如果我有时间，我会更深入地了解GNU并行，但这对我自己的“doCalculations”过程来说很困难。

简单地说:

#!/bin/bash

pids=""

for i in `seq 0 9`; do
   doCalculations $i &
   pids="$pids $!"
done

wait $pids

...code continued here ...

更新:

正如许多评论者指出的那样，上面的代码在继续之前会等待所有进程完成，但如果其中一个进程失败了，它不会退出，也不会失败，可以通过@Bryan， @SamBrightman和其他人建议的以下修改来完成:

#!/bin/bash

pids=""
RESULT=0


for i in `seq 0 9`; do
   doCalculations $i &
   pids="$pids $!"
done

for pid in $pids; do
    wait $pid || let "RESULT=1"
done

if [ "$RESULT" == "1" ];
    then
       exit 1
fi

...code continued here ...