在Bash中给定超时后如何杀死子进程? [英] How to kill a child process after a given timeout in Bash?

问题描述

我有一个bash脚本，该脚本启动一个子进程，该子进程有时会崩溃(实际上是挂起)，并且没有明显的原因(封闭源代码，因此我无能为力).结果，我希望能够在给定的时间内启动此过程，并在给定的时间之后未成功返回的情况下将其终止.

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.

是否有使用bash实现此目标的简单和鲁棒方法?

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

P.S .:告诉我这个问题是否更适合于serverfault或超级用户.

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

推荐答案

(如下所示: BASH FAQ条目#68:我如何运行命令，并在N秒后中止(超时)? ")

(As seen in: BASH FAQ entry #68: "How do I run a command, and have it abort (timeout) after N seconds?")

如果您不介意下载某些内容，请使用timeout(sudo apt-get install timeout)并按以下方式使用它:(大多数系统已安装，否则请使用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，请执行内部超时操作:

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代码超时，请使用第二个选项，例如:

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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