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

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。

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

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

http://jeremy.zawodny.com/blog/archives/010717.html:

#!/bin/bash

FAIL=0

echo "starting"

./sleeper 2 0 &
./sleeper 2 1 &
./sleeper 3 0 &
./sleeper 2 0 &

for job in `jobs -p`
do
echo $job
    wait $job || let "FAIL+=1"
done

echo $FAIL

if [ "$FAIL" == "0" ];
then
echo "YAY!"
else
echo "FAIL! ($FAIL)"
fi

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。