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?
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)