如何在 Bash 中给定超时后杀死子进程?

问题描述

我有一个 bash 脚本，它启动一个子进程，该进程不时崩溃(实际上，挂起)并且没有明显的原因(封闭源代码，所以我无能为力).因此，我希望能够在给定的时间内启动此进程，如果在给定的时间后没有成功返回，则将其终止.

I have a bash script that launches a child process that crashes (actually, hangs) from time to time and with no apparent reason (closed source, so there isn't much I can do about it). As a result, I would like to be able to launch this process for a given amount of time, and kill it if it did not return successfully after a given amount of time.

是否有一种简单和强大的方法来使用 bash 实现这一目标?

Is there a simple and robust way to achieve that using bash?

P.S.:告诉我这个问题更适合 serverfault 还是超级用户.

P.S.: tell me if this question is better suited to serverfault or superuser.

推荐答案

(见:BASH 常见问题条目 #68:如何运行命令，并让它在 N 秒后中止(超时)?")

如果您不介意下载某些东西，请使用 timeout (sudo apt-get install timeout) 并使用它:(大多数系统已经安装了它，否则使用sudo apt-get install coreutils)

If you don't mind downloading something, use timeout (sudo apt-get install timeout) and use it like: (most Systems have it already installed otherwise use sudo apt-get install coreutils)

timeout 10 ping www.goooooogle.com

如果您不想下载某些内容，请执行 timeout 在内部执行的操作:

If you don't want to download something, do what timeout does internally:

( cmdpid=$BASHPID; (sleep 10; kill $cmdpid) & exec ping www.goooooogle.com )

如果您想为更长的 bash 代码设置超时，请使用第二个选项:

In case that you want to do a timeout for longer bash code, use the second option as such:

( cmdpid=$BASHPID;
    (sleep 10; kill $cmdpid)
   & while ! ping -w 1 www.goooooogle.com
     do
         echo crap;
     done )

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