等待所有进程具有特定名称来完成（在bash） [英] Wait for all processes with a certain name to finish (in bash)

问题描述

我想在linux下等（Ubuntu的11.10）的过程来完成的负载。每个这些方法具有不同的PID，但相同的名称。是否有可能做到这一点？

编辑：

也许我应该指定一个我并不知道什么是PID，只是过程的名称。

编辑：

感谢您的答案;凯文似乎做我想做的。但是，它不会在具体的应用我有工作，所以我已经张贴了更详细的后续问题，<一个href=\"http://stackoverflow.com/questions/8040417/parrallel-processing-in-shell-scripting-pid-is-not-a-child-of-this-shell\">here.

解决方案

 等待$（指派，PROGRAMNAME）
 

应该做的。

I would like to wait in linux (Ubuntu 11.10) for a load of processes to finish. Each of these processes has a different pid but the same name. Is it possible to do this?

EDIT:

Perhaps I should specify that I don't necessarily know what the pid are, just the process name.

EDIT:

Thanks for the answers; Kevin's seems to do what I want. However, it doesn't work in the specific application I have, so I've posted a more detailed follow-up question here.

解决方案

wait $(pgrep programName)

Ought to do it.

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