Python AttributeError: 'Flask' 对象没有属性 'user_options'
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AttributeError: 'Flask' object has no attribute 'user_options'
提问by Ravdeep
I am trying to setup this basic example from the following doc:
我正在尝试从以下文档中设置这个基本示例：
But so far I keep getting the below error:
但到目前为止，我一直收到以下错误：
AttributeError: 'Flask' object has no attribute 'user_options'
AttributeError: 'Flask' 对象没有属性 'user_options'
I am using celery 3.1.15.
我正在使用芹菜 3.1.15。
from celery import Celery
def make_celery(app):
celery = Celery(app.import_name, broker=app.config['CELERY_BROKER_URL'])
celery.conf.update(app.config)
TaskBase = celery.Task
class ContextTask(TaskBase):
abstract = True
def __call__(self, *args, **kwargs):
with app.app_context():
return TaskBase.__call__(self, *args, **kwargs)
celery.Task = ContextTask
return celery
Example:
例子：
from flask import Flask
app = Flask(__name__)
app.config.update(
CELERY_BROKER_URL='redis://localhost:6379',
CELERY_RESULT_BACKEND='redis://localhost:6379'
)
celery = make_celery(app)
@celery.task()
def add_together(a, b):
return a + b
Traceback error:
追溯错误：
Traceback (most recent call last):
File "/usr/local/bin/celery", line 11, in <module>
sys.exit(main())
File "/usr/local/lib/python2.7/dist-packages/celery/__main__.py", line 30, in main
main()
File "/usr/local/lib/python2.7/dist-packages/celery/bin/celery.py", line 81, in main
cmd.execute_from_commandline(argv)
File "/usr/local/lib/python2.7/dist-packages/celery/bin/celery.py", line 769, in execute_from_commandline
super(CeleryCommand, self).execute_from_commandline(argv)))
File "/usr/local/lib/python2.7/dist-packages/celery/bin/base.py", line 305, in execute_from_commandline
argv = self.setup_app_from_commandline(argv)
File "/usr/local/lib/python2.7/dist-packages/celery/bin/base.py", line 473, in setup_app_from_commandline
user_preload = tuple(self.app.user_options['preload'] or ())
AttributeError: 'Flask' object has no attribute 'user_options'
采纳答案by TomL
The Flask Celery Based Background Tasks page (http://flask.pocoo.org/docs/patterns/celery/) suggests this to start celery:
基于 Flask Celery 的后台任务页面 ( http://flask.pocoo.org/docs/patterns/celery/) 建议这样启动 celery：
celery -A your_application worker
celery -A your_application worker
The your_application string has to point to your application's package or module that creates the celery object.
your_application 字符串必须指向创建 celery 对象的应用程序包或模块。
Assuming the code resides in application.py, explicitly pointing to the celery object (not just the module name) avoided the error:
假设代码驻留在 application.py 中，显式指向 celery 对象（不仅仅是模块名称）避免了错误：
celery -A application.celery worker
celery -A application.celery worker
回答by nhywieza
rename app flask_app It will work
重命名应用程序flask_app 它将起作用
回答by user7374381
like this:
像这样：
celery -A your_application worker
where your_application stands:
your_application 所在的位置：
your_application = Flask(\__name\__)
the python file name: your_application.py, it will work
python 文件名：your_application.py，它将起作用
By the way, celery v4 is unsupported in Windows
顺便说一句，芹菜 v4 在 Windows 中不受支持
回答by sri
This worked for me:
这对我有用：
celery -A my_app_module_name.celery worker
