Linux 如何在 Bash 中运行超时的进程？
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How to run a process with a timeout in Bash?
提问by AlanF
Possible Duplicate:
Bash script that kills a child process after a given timeout
可能的重复：
在给定超时后杀死子进程的 Bash 脚本
Is there a way to write a shell script that would execute a certain command for 15 seconds, then kill the command?
有没有办法编写一个 shell 脚本来执行某个命令 15 秒，然后终止该命令？
I have tried sleep, wait and ping but maybe I am using them wrong.
我试过 sleep、wait 和 ping，但也许我用错了。
回答by ArjunShankar
Some machines don't have timeoutinstalled/available. In that case, you could background the process; its PID then gets stored as $!; then sleep for the required amount of time, then kill it:
有些机器没有timeout安装/可用。在这种情况下，你可以后台处理；它的PID然后被存储为$!; 然后睡眠所需的时间，然后杀死它：
some_command arg1 arg2 &
TASK_PID=$!
sleep 15
kill $TASK_PID
At this URLI find that there are mentioned, more than one solutions to make this happen.
在这个 URL 上，我发现提到了不止一种解决方案来实现这一目标。
回答by Karoly Horvath
Use the timeoutcommand:
使用timeout命令：
timeout 15s command
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 timeoutscript or small C app that does the same thing.
注意：在某些系统上您需要安装coreutils，在其他系统上它丢失或具有不同的命令行参数。请参阅@ArjunShankar 发布的替代解决方案。基于它，您可以封装样板代码并创建您自己的可移植timeout脚本或执行相同操作的小型 C 应用程序。
