Shell脚本在延迟后产生一个进程 [英] Shell script spawning a process after a delay

问题描述

在Shell脚本中出现延迟后，如何生成一个进程?我希望命令在脚本启动后60秒启动，但是我希望继续运行脚本的其余部分，而不必先等待60秒.这是主意:

How can I spawn a process after a delay in a shell script? I want a command to start 60 seconds after the script starts, but I want to keep running the rest of the script without waiting 60 seconds first. Here's the idea:

#!/bin/sh
# Echo A 60 seconds later, but without blocking the rest of the script
sleep 60 && echo "A"

echo "B"
echo "C"

输出应为

B
C
... 60 seconds later
A

我需要能够在一个脚本中完成所有这一切.IE.没有创建从第一个外壳程序脚本调用的第二个脚本.

I need to be able to do this all in one script. Ie. no creating a second script that is called from the first shell script.

推荐答案

&开始后台工作，所以

& starts a background job, so

sleep 60 && echo "A" &

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