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

Question

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?

Answer 1

Based on code from Django-Celery module you could try something like this:

from __future__ import absolute_import, unicode_literals

from celery import current_app
from celery.bin import worker


if __name__ == '__main__':
    app = current_app._get_current_object()

    worker = worker.worker(app=app)

    options = {
        'broker': 'amqp://guest:guest@localhost:5672//',
        'loglevel': 'INFO',
        'traceback': True,
    }

    worker.run(**options)

Answer 2

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)

Answer 3

I think you are just missing wrapping the args so celery can read them, like:

queue = Celery('blah', include=['blah'])
queue.start(argv=['celery', 'worker', '-l', 'info'])