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。

正是为了这个目的，我写了一个bash函数:for。

注意::for不仅保留并返回失败函数的退出码，而且终止所有并行运行的实例。在这种情况下可能不需要。

#!/usr/bin/env bash

# Wait for pids to terminate. If one pid exits with
# a non zero exit code, send the TERM signal to all
# processes and retain that exit code
#
# usage:
# :wait 123 32
function :wait(){
    local pids=("$@")
    [ ${#pids} -eq 0 ] && return $?

    trap 'kill -INT "${pids[@]}" &>/dev/null || true; trap - INT' INT
    trap 'kill -TERM "${pids[@]}" &>/dev/null || true; trap - RETURN TERM' RETURN TERM

    for pid in "${pids[@]}"; do
        wait "${pid}" || return $?
    done

    trap - INT RETURN TERM
}

# Run a function in parallel for each argument.
# Stop all instances if one exits with a non zero
# exit code
#
# usage:
# :for func 1 2 3
#
# env:
# FOR_PARALLEL: Max functions running in parallel
function :for(){
    local f="${1}" && shift

    local i=0
    local pids=()
    for arg in "$@"; do
        ( ${f} "${arg}" ) &
        pids+=("$!")
        if [ ! -z ${FOR_PARALLEL+x} ]; then
            (( i=(i+1)%${FOR_PARALLEL} ))
            if (( i==0 )) ;then
                :wait "${pids[@]}" || return $?
                pids=()
            fi
        fi
    done && [ ${#pids} -eq 0 ] || :wait "${pids[@]}" || return $?
}

使用

for.sh:

#!/usr/bin/env bash
set -e

# import :for from gist: https://gist.github.com/Enteee/c8c11d46a95568be4d331ba58a702b62#file-for
# if you don't like curl imports, source the actual file here.
source <(curl -Ls https://gist.githubusercontent.com/Enteee/c8c11d46a95568be4d331ba58a702b62/raw/)

msg="You should see this three times"

:(){
  i="${1}" && shift

  echo "${msg}"

  sleep 1
  if   [ "$i" == "1" ]; then sleep 1
  elif [ "$i" == "2" ]; then false
  elif [ "$i" == "3" ]; then
    sleep 3
    echo "You should never see this"
  fi
} && :for : 1 2 3 || exit $?

echo "You should never see this"

$ ./for.sh; echo $?
You should see this three times
You should see this three times
You should see this three times
1

参考文献

[1]:博客 [2]:要点

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

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

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