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, Sep 17 '14 at 07:50
I am not configuring anything on celery side.I am just following that documentation directly. — Ravdeep, Sep 17 '14 at 08:00
Guys, thanks for your help. Fixed the issue. I was running the celery worker incorrectly.Silly mistake on my side. — Ravdeep, Sep 18 '14 at 09:35

score 115 · Accepted Answer · edited Jul 12 '18 at 08:44

115

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 '18 at 08:44

Mel

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answered Sep 24 '14 at 16:25

TomL

2,724
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14

I've been running into a number of issues of this kind. Celery's a bit magical for my tastes. – Eric Walker Oct 11 '14 at 21:07
7

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 – Calumah May 07 '18 at 17:38
This worked for me ️ – Miebaka Feb 28 '22 at 10:12
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. – nicorellius May 01 '22 at 19:21

score 8 · Answer 2 · answered Aug 08 '18 at 13:00

8

This worked for me:

celery -A my_app_module_name.celery worker

answered Aug 08 '18 at 13:00

sri

744
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For me as well. I use `celery 4.3.0 (rhubarb)` – Martin Thoma Jun 10 '19 at 05:09

score 0 · Answer 3 · answered Apr 26 '16 at 09:31

0

rename app flask_app It will work

answered Apr 26 '16 at 09:31

nhywieza

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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). – Reti43 Apr 26 '16 at 21:20
@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. – Jainabhi May 30 '22 at 08:51

score -4 · Answer 4 · edited Jan 04 '17 at 15:44

-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 04 '17 at 15:44

Caleb Kleveter

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answered Jan 04 '17 at 14:22

user7374381

9

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

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