AttributeError: 'Flask' object has no attribute 'user_options'

Question

I am trying to setup this basic example from the following doc:

http://flask.pocoo.org/docs/patterns/celery/

But so far I keep getting the below error:

AttributeError: 'Flask' object has no attribute 'user_options'

I am using celery 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:

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:

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'

What command line did you use to run Celery? How did you configure Celery itself? You only showed us the Flask integration side. — Martijn Pieters
– Martijn Pieters, Commented Sep 17, 2014 at 7:50
so I am using "celery -A <my_application> worker" to run the celery. — Ravdeep
– Ravdeep, Commented Sep 17, 2014 at 7:52
I am not configuring anything on celery side.I am just following that documentation directly. — Ravdeep
– Ravdeep, Commented Sep 17, 2014 at 8:00

Mel · Accepted Answer · 2018-07-12 08:44:17Z

128

The Flask Celery Based Background Tasks page (http://flask.pocoo.org/docs/patterns/celery/) suggests this to start celery:

celery -A your_application worker

The your_application string has to point to your application’s package or module that creates the celery object.

Assuming the code resides in application.py, explicitly pointing to the celery object (not just the module name) avoided the error:

celery -A application.celery worker

edited Jul 12, 2018 at 8:44

Mel

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answered Sep 24, 2014 at 16:25

TomL

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5 Comments

Eric Walker Over a year ago

I've been running into a number of issues of this kind. Celery's a bit magical for my tastes.

Calumah Over a year ago

I got same problem today... My issue was app = Flask(__name__) because it clash with celery internal self.app. So you just need to rename app = Flask(__name__) to "application" or something else to avoid this traceback

Miebaka Over a year ago

This worked for me 🚀️

nicorellius Over a year ago

I love Celery but the docs leave MUCH to be desired. I suppose they only cater to veteran network and web programmers... Kinda hostile toward newbs.

fudu Over a year ago

For me, I was following this instruction link for celery on Flask: flask.palletsprojects.com/en/3.0.x/patterns/celery so I changed from celery -A example worker to celery -A example.celery_app worker to make it work.

sri · Accepted Answer · 2018-08-08 13:00:13Z

11

This worked for me:

celery -A my_app_module_name.celery worker

answered Aug 8, 2018 at 13:00

sri

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1 Comment

Martin Thoma Over a year ago

For me as well. I use celery 4.3.0 (rhubarb)

nhywieza · Accepted Answer · 2016-04-26 09:31:00Z

0

rename app flask_app It will work

answered Apr 26, 2016 at 9:31

nhywieza

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2 Comments

Reti43 Over a year ago

This is irrelevant. The OP is not struggling with any NameError and your answer is quite vague how it could solve the problem (which it seems it doesn't anyway).

Jainabhi Over a year ago

@Reti43 if the file name is app.py, then app = Flask(__name__) will have a conflict. To resolve this, he suggested renaming the app to flask_app. I agree that @nhywieza did not give the proper context.

akpp · Accepted Answer · 2024-12-09 21:29:10Z

0

Celery is trying to run wrong app it's Flask instance
solotion1: rename app -> flask_app or other name
solution2: specify Celery instance celery -A my_app_module_name.celery worker (added .celery)

answered Dec 9, 2024 at 21:29

akpp

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Comments

Caleb Kleveter · Accepted Answer · 2017-01-04 15:44:13Z

-4

like this:

celery -A your_application worker

where your_application stands:

your_application = Flask(\__name\__)

the python file name: your_application.py, it will work

By the way, celery v4 is unsupported in Windows

edited Jan 4, 2017 at 15:44

Caleb Kleveter

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answered Jan 4, 2017 at 14:22

user7374381

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