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
465 views
in Technique[技术] by (71.8m points)

python - How to start a Celery worker from a script/module __main__?

I've define a Celery app in a module, and now I want to start the worker from the same module in its __main__, i.e. by running the module with python -m instead of celery from the command line. I tried this:

app = Celery('project', include=['project.tasks'])

# do all kind of project-specific configuration
# that should occur whenever this module is imported

if __name__ == '__main__':
    # log stuff about the configuration
    app.start(['worker', '-A', 'project.tasks'])

but now Celery thinks I'm running the worker without arguments:

Usage: worker <command> [options] 

Show help screen and exit.

Options:
  -A APP, --app=APP     app instance to use (e.g. module.attr_name)
[snip]

The usage message is the one you get from celery --help, as if it didn't get a command. I've also tried

app.worker_main(['-A', 'project.tasks'])

but that complains about the -A not being recognized.

So how do I do this? Or alternatively, how do I pass a callback to the worker to have it log information about its configuration?

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)

using app.worker_main method (v3.1.12):

± cat start_celery.py
#!/usr/bin/python

from myapp import app


if __name__ == "__main__":
    argv = [
        'worker',
        '--loglevel=DEBUG',
    ]
    app.worker_main(argv)

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

...