等待进程直到所有子进程完成?

wait process until all subprocess finish?(等待进程直到所有子进程完成?)

问题描述

我有一个创建两个或多个子进程的主进程，我希望主进程等到所有子进程完成操作并退出?

I have a main process which creates two or more sub processes, I want main process to wait until all sub processes finish their operations and exits?

 # main_script.py

 p1 = subprocess.Popen(['python script1.py']) 
 p2 = subprocess.Popen(['python script2.py'])
 ... 
 #wait main process until both p1, p2 finish
 ...

推荐答案

一个 Popen 对象有一个 .wait() 方法正是为此定义的:等待给定子进程的完成(此外，对于重新调整其退出状态).

A Popen object has a .wait() method exactly defined for this: to wait for the completion of a given subprocess (and, besides, for retuning its exit status).

如果你使用这种方法，你可以防止进程僵尸停留太久.

If you use this method, you'll prevent that the process zombies are lying around for too long.

(或者，您可以使用 subprocess.call() 或 subprocess.check_call() 用于调用和等待.如果您不需要进程的 IO，那可能就足够了.但这可能不是一个选项，因为您的 if 两个子进程似乎应该在其中运行并行，他们不会使用 (check_)call().)

(Alternatively, you can use subprocess.call() or subprocess.check_call() for calling and waiting. If you don't need IO with the process, that might be enough. But probably this is not an option, because your if the two subprocesses seem to be supposed to run in parallel, which they won't with (check_)call().)

如果你有几个子流程要等待，你可以这样做

If you have several subprocesses to wait for, you can do

exit_codes = [p.wait() for p in p1, p2]

所有子进程完成后立即返回.然后，您将获得一个返回代码列表，您可以对其进行评估.

which returns as soon as all subprocesses have finished. You then have a list of return codes which you maybe can evaluate.

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