Activity

1. jezdez

   changed the title [-]Add getting started section[/-] on Mar 2, 2018
2. timorthi commented on Mar 2, 2018

   

   timorthi

   on Mar 2, 2018

   More actions

   The current README is a good start, it's just missing a simple example. Maybe adding a section like this would be sufficient?

   ---

   Example Usage

   Define a job and queue it:

   # app.py
   
   from flask import Flask
   from flask_rq2 import RQ
   
   app = Flask(__name__)
   app.config['RQ_REDIS_URL'] = 'redis://127.0.0.1:6379/0'
   
   rq = RQ(app)
   
   
   @rq.job 
   def foo():
       some_long_running_task()
   
   
   @app.route('/')
   def index():
       foo.queue()
       return 'Job queued!'
   

   Start a worker process:

   $ export FLASK_APP=app.py # Required for jobs to run within the Flask app context
   $ flask rq worker
   

   And you're ready to go!

   👍React with 👍1Reacted by easyandme

   ❤️React with ❤️1Reacted by easyandme
3. easyandme commented on Mar 10, 2018

   

   easyandme

   on Mar 10, 2018

   More actions

   @timorthi I got
   AttributeError: module '__main__' has no attribute 'add'
   Any thoughts? Many thanks!
4. timorthi commented on Mar 10, 2018

   

   timorthi

   on Mar 10, 2018

   More actions

   Hey @maxim-xu, if you're following the example I posted above, there's no add() function defined anywhere, only foo(). Other than that I don't think I can help you without looking at your code.
5. easyandme commented on Mar 10, 2018

   

   easyandme

   on Mar 10, 2018

   More actions

   @timorthi Sorry! I'd appreciate it if you could take a look at this.
   I just replaced foo() with add. The rest of the code is almost the same, except that at the bottom I have:

   if __name__ == "__main__": 
       port = int(os.environ.get("PORT", 5000))
       application.run(port=port, debug=False)
   

   because I was running the flask server locally.
   When I run: python app.py & flask rq worker it shows the error I mentioned above.
6. timorthi commented on Mar 10, 2018

   

   timorthi

   on Mar 10, 2018

   Last edited by timorthi

   More actions

   @maxim-xu Ah ok, this seems to be happening because you're using the Flask.run() command instead of the CLI. In your case the app's __name__ is __main__, which is why the worker process is trying to resolve __main__.foo(). I think you could get around this by naming your file __main__.py, but that's probably not recommended.

   I'm not too well-versed with how Flask CLI or RQ handles __name__ with regard to the app instance so I can't comment further, but if you start your app using the Flask CLI FLASK_APP=app.py flask run then it'll solve your problem. Initializing your app instance with something like application = Flask('app') (assuming your file name is app.py) should do the trick too.
7. easyandme commented on Mar 10, 2018

   

   easyandme

   on Mar 10, 2018

   More actions

   @timorthi Solved. Thank you so much!

   🎉React with 🎉1Reacted by Timothy Ng
8. blacksteel1288 commented on Aug 28, 2023

   

   blacksteel1288

   on Aug 28, 2023

   More actions

   My app runs fine, but when I do a 'flask rq worker' on the CLI to start the worker process I'm getting:

   "Error: Detected factory 'create_app' in module 'app', but could not call it without arguments. Use 'app:create_app(args)' to specify arguments."

   I'm using an application factory pattern with blueprints.

   Any thoughts?