当任何子进程以代码 !=0 结束时，如何在 bash 中等待多个子进程完成并返回退出代码 !=0? [英] How to wait in bash for several subprocesses to finish and return exit code !=0 when any subprocess ends with code !=0?

问题描述

如何在 bash 脚本中等待从该脚本产生的多个子进程完成并在任何子进程以代码 !=0 结束时返回退出代码 !=0 ?

How to wait in a bash script for several subprocesses spawned from that script to finish and return exit code !=0 when any of the subprocesses ends with code !=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 个生成的子进程，但它总是给出退出状态 0(参见 help wait).如何修改此脚本，以便在任何子进程以代码 !=0 结束时发现生成的子进程的退出状态并返回退出代码 1?

The above script will wait for all 10 spawned subprocesses, but it will always give exit status 0 (see help wait). How can I modify this script so it will discover exit statuses of spawned subprocesses and return exit code 1 when any of subprocesses ends with code !=0?

有没有比收集子进程的PID，按顺序等待它们并总结退出状态更好的解决方案?

Is there any better solution for that than collecting PIDs of the subprocesses, wait for them in order and sum exit statuses?

推荐答案

wait 还(可选)获取要等待的进程的 PID，并使用 $! 你会得到在后台启动的最后一个命令的 PID.修改循环，将每个生成的子进程的 PID 存储到一个数组中，然后再次循环等待每个 PID.

wait also (optionally) takes the PID of the process to wait for, and with $! you get the PID of the last command launched in the background. Modify the loop to store the PID of each spawned sub-process into an array, and then loop again waiting on each PID.

# run processes and store pids in array
for i in $n_procs; do
    ./procs[${i}] &
    pids[${i}]=$!
done

# wait for all pids
for pid in ${pids[*]}; do
    wait $pid
done

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