使 bash wait 命令在第一个错误时失败

Question

我有一个 bash 脚本，可以并行执行一些耗时的命令，到目前为止它运行得很好。我正在使用等待命令，如下所示：

docker pull */* &
docker pull */* &
docker pull */* &
docker pull */* &
docker pull */* &
docker pull */* &
docker pull */* &
composer install -n &
wait

现在，我希望此脚本中止所有命令，并在其中一个命令失败时给出退出代码。如何实现这一目标？

注：*/*是docker镜像名称，对于上下文来说不重要

Answer 1

这将需要 bash wait -n。

这里的技巧是保留您生成的子进程的列表，然后单独等待它们（按照它们完成的顺序）。然后，您可以检查已完成进程的返回代码，如果失败则杀死其中的很多进程。例如：

#!/bin/bash

# Remember the pid after each command, keep a list of them.
# pidlist=foo could also go on a line of its own, but I
# find this more readable because I like tabular layouts.
sleep 10 & pidlist="$!"
sleep 10 & pidlist="$pidlist $!"
sleep 10 & pidlist="$pidlist $!"
sleep 10 & pidlist="$pidlist $!"
sleep 10 & pidlist="$pidlist $!"
sleep 10 & pidlist="$pidlist $!"
false    & pidlist="$pidlist $!"

echo $pidlist

# $pidlist intentionally unquoted so every pid in it expands to a
# parameter of its own. Note that $i doesn't have to be the PID of
# the process that finished, it's just important that the loop runs
# as many times as there are PIDs.
for i in $pidlist; do
    # Wait for a single process in the pid list to finish,
    # check if it failed,
    if ! wait -n $pidlist; then
        # If it did, kill the lot. pipe kill's stderr
        # to /dev/null so it doesn't complain about
        # already-finished processes not existing
        # anymore.
        kill $pidlist 2>/dev/null

        # then exit with a non-zero status.
        exit 1
    fi
done