我在模块中定义了一个Celery
应用,现在我想从__main__
中的同一个模块启动该工作,即通过{{1}运行模块而不是命令行中的python -m
。我试过这个:
celery
但是现在Celery认为我没有参数运行工人:
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'])
使用消息是您从Usage: worker <command> [options]
Show help screen and exit.
Options:
-A APP, --app=APP app instance to use (e.g. module.attr_name)
[snip]
获得的消息,就好像它没有获得命令一样。我也试过了
celery --help
但抱怨app.worker_main(['-A', 'project.tasks'])
无法识别。
那我该怎么做?或者,如何将回调传递给worker以使其记录有关其配置的信息?
答案 0 :(得分:17)
使用app.worker_main方法(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)
答案 1 :(得分:10)
基于code from Django-Celery module你可以尝试这样的事情:
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)
答案 2 :(得分:3)
我认为你只是缺少包裹args所以芹菜可以读取它们,如:
queue = Celery('blah', include=['blah'])
queue.start(argv=['celery', 'worker', '-l', 'info'])
答案 3 :(得分:1)
worker_main
的结果如下:
AttributeError: 'Celery' object has no attribute 'worker_main'
app = celery.Celery(
'project',
include=['project.tasks']
)
if __name__ == '__main__':
worker = app.Worker(
include=['project.tasks']
)
worker.start()
有关详细信息,请参见此处celery.apps.worker和celery.worker.WorkController.setup_defaults(希望将来会得到更好的记录)。
答案 4 :(得分:1)
worker_main
被放回芹菜5.0.3中:
https://github.com/celery/celery/pull/6481
这对我而言适用于5.0.4:
self.app.worker_main(argv = ['worker', '--loglevel=info', '--concurrency={}'.format(os.environ['CELERY_CONCURRENCY']), '--without-gossip'])