如何从脚本/模块__main__启动Celery worker？ [英] How to start a Celery worker from a script/module __main__?

问题描述

我已经在一个模块中定义了一个 Celery 应用程序，现在我想从其 __ main__中的同一模块启动worker。 code>，即通过在命令行中使用 python -m 而不是 celery 运行模块。我尝试过这个：

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'])

但现在Celery认为我正在运行没有参数的工作器：

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]

用法消息是您从 celery --help 收到的消息，好像没有没有命令。我也尝试过

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'])

但这抱怨 -A 无法被识别。

那么我该怎么做？或者，如何将回调传递给工作程序以使其记录有关其配置的信息？

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?

推荐答案

基于来自Django-Celery模块的代码像这样的东西：

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)

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