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

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。

其他回答

这是我使用的东西:

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

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}

我刚刚修改了一个脚本到后台和并行化的过程。

我做了一些实验(在Solaris上使用bash和ksh),发现如果退出状态不为零,'wait'将输出退出状态,或者当没有提供PID参数时,将输出一个返回非零退出的作业列表。如。

Bash:

$ sleep 20 && exit 1 &
$ sleep 10 && exit 2 &
$ wait
[1]-  Exit 2                  sleep 20 && exit 2
[2]+  Exit 1                  sleep 10 && exit 1

Ksh:

$ sleep 20 && exit 1 &
$ sleep 10 && exit 2 &
$ wait
[1]+  Done(2)                  sleep 20 && exit 2
[2]+  Done(1)                  sleep 10 && exit 1

这个输出被写入stderr,所以OPs示例的简单解决方案可以是:

#!/bin/bash

trap "rm -f /tmp/x.$$" EXIT

for i in `seq 0 9`; do
  doCalculations $i &
done

wait 2> /tmp/x.$$
if [ `wc -l /tmp/x.$$` -gt 0 ] ; then
  exit 1
fi

虽然这:

wait 2> >(wc -l)

也将返回一个计数,但不包含TMP文件。这也可以这样使用,例如:

wait 2> >(if [ `wc -l` -gt 0 ] ; then echo "ERROR"; fi)

但是这并不比tmp文件有用多少。我找不到一种有效的方法来避免tmp文件,同时也避免在子shell中运行“等待”,这根本不会起作用。

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