如何获取bash中多个并行后台进程的退出状态 [英] How to get the exit status of multiple parallel background processes in bash

问题描述

在Bash中，我想并行调用多个命令，同时捕获所有进程的退出代码.

In Bash, I want to call a command multiple times in parallel, while capturing all the processes exit codes.

我知道如何启动并等待它们，但是等待只会给我最后退出的进程的退出代码.我还需要寿命较短的退出代码.

I know how to start and wait for them, but wait will only give me the exit code of the last process exiting. I also need the exit code of the shorter lived processes.

不幸的是，我没有bash 4.3，所以 wait -n 不是一个选项，也不像

Unfortunately I don't have bash 4.3, so wait -n is not an option, nor is gnu parallel as suggested in #3004811

#16032001 几乎问了同样的问题，但那里也没有提供解决方案.

#16032001 pretty much asks the same question but no solution was offered there either.

我目前唯一想到的方法是编写一个帮助程序脚本，将退出代码存储在文件中，但这听起来并不干净.

The only way I can currently think of is writing a helper script that stores the exit codes in a file, but this doesn't sound like a clean solution.

推荐答案

答案在我没有意识到，尽管孩子立即被bash收割，但是内置的等待仍然可以访问该pid的退出代码.

I was unaware that though the child is immediately reaped by bash, the builtin wait can still access the exit code for the pid.

#!/bin/bash

FAIL=0
PIDS=""

echo "starting"

sleep 5 &
PIDS="$PIDS $!"

sleep 3 &
PIDS="$PIDS $!"

/bin/false &
PIDS="$PIDS $!"

sleep 3 &
PIDS="$PIDS $!"

for job in $PIDS
do
    wait $job || let "FAIL+=1"
    echo $job $FAIL
done

echo $FAIL

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

正确给出

starting
14772 0
14773 0
14774 1
14775 1
1
FAIL! (1)

只有第三个进程(/bin/false)失败，这在第三行中从0到1的切换表示.

Only the third process (/bin/false) fails, indicated by the switch from 0 to one in the third line.

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