Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
246 views
in Technique[技术] by (71.8m points)

python - How to start a long-running process from a Django view?

I need to run a process that might take hours to complete from a Django view. I don't need to know the state or communicate with it but I need that view to redirect away right after starting the process.

I've tried using subprocess.Popen, using it within a new threading.Thread, multiprocessing.Process. However, the parent process keeps hanging until child terminates. The only way that almost gets it done is using a fork. Obviously that isn't good as it leaves a zombie process behind until parent terminates.

That's what I'm trying to do when using fork:

if os.fork() == 0:
    subprocess.Popen(["/usr/bin/python", script_path, "-v"])
else:
    return HttpResponseRedirect(reverse('view_to_redirect'))

So, is there a way to run a completely independent process from a Django view with minimal casualties? Or am I doing something wrong?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I don't know if this will be suitable for your case, nevertheless here is what I do: I use a task queue (via a django model); when the view is called, it enters a new record in the tasks and redirects happily. Tasks in turn are executed by cron on a regular basis independently from django.

Edit: cron calls the relevant (and custom) django command to execute the task.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...