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

这是有效的,应该是一个很好的,如果不是更好的@HoverHell的答案!

#!/usr/bin/env bash

set -m # allow for job control
EXIT_CODE=0;  # exit code of overall script

function foo() {
     echo "CHLD exit code is $1"
     echo "CHLD pid is $2"
     echo $(jobs -l)

     for job in `jobs -p`; do
         echo "PID => ${job}"
         wait ${job} ||  echo "At least one test failed with exit code => $?" ; EXIT_CODE=1
     done
}

trap 'foo $? $$' CHLD

DIRN=$(dirname "$0");

commands=(
    "{ echo "foo" && exit 4; }"
    "{ echo "bar" && exit 3; }"
    "{ echo "baz" && exit 5; }"
)

clen=`expr "${#commands[@]}" - 1` # get length of commands - 1

for i in `seq 0 "$clen"`; do
    (echo "${commands[$i]}" | bash) &   # run the command via bash in subshell
    echo "$i ith command has been issued as a background job"
done

# wait for all to finish
wait;

echo "EXIT_CODE => $EXIT_CODE"
exit "$EXIT_CODE"

# end

当然,我已经在一个NPM项目中保存了这个脚本,它允许你并行运行bash命令,对测试很有用:

https://github.com/ORESoftware/generic-subshell

其他回答

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

捕获CHLD信号可能不起作用,因为如果它们同时到达,您可能会丢失一些信号。

#!/bin/bash

trap 'rm -f $tmpfile' EXIT

tmpfile=$(mktemp)

doCalculations() {
    echo start job $i...
    sleep $((RANDOM % 5)) 
    echo ...end job $i
    exit $((RANDOM % 10))
}

number_of_jobs=10

for i in $( seq 1 $number_of_jobs )
do
    ( trap "echo job$i : exit value : \$? >> $tmpfile" EXIT; doCalculations ) &
done

wait 

i=0
while read res; do
    echo "$res"
    let i++
done < "$tmpfile"

echo $i jobs done !!!

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。

我看到这里列出了很多很好的例子,我也想把我的举出来。

#! /bin/bash

items="1 2 3 4 5 6"
pids=""

for item in $items; do
    sleep $item &
    pids+="$! "
done

for pid in $pids; do
    wait $pid
    if [ $? -eq 0 ]; then
        echo "SUCCESS - Job $pid exited with a status of $?"
    else
        echo "FAILED - Job $pid exited with a status of $?"
    fi
done

我使用非常类似的方法并行启动/停止服务器/服务,并检查每个退出状态。对我来说很好。希望这能帮助到一些人!

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