如何在Bash中运行带有超时的进程? [英] How to run a process with a timeout in Bash?

问题描述

可能重复:
在给定时间后杀死子进程的Bash脚本超时

Possible Duplicate:
Bash script that kills a child process after a given timeout

有没有一种方法可以编写将执行某条命令15秒钟然后杀死该命令的Shell脚本?

Is there a way to write a shell script that would execute a certain command for 15 seconds, then kill the command?

我尝试了睡眠，等待和ping操作，但也许我在错误地使用它们.

I have tried sleep, wait and ping but maybe I am using them wrong.

推荐答案

使用timeout命令:

timeout 15s command

注意:在某些系统上，您需要安装coreutils，在其他系统上，则需要安装coreutils或具有不同的命令行参数.请参阅@ArjunShankar发布的替代解决方案.在此基础上，您可以封装样板代码，并创建自己的可移植timeout脚本或执行相同操作的小型C应用程序.

Note: on some systems you need to install coreutils, on others it's missing or has different command line arguments. See an alternate solution posted by @ArjunShankar . Based on it you can encapsulate that boiler-plate code and create your own portable timeout script or small C app that does the same thing.

这篇关于如何在Bash中运行带有超时的进程?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！