Why does running the Flask dev server run itself twice?

Question

I\'m using Flask for developing a website and while in development I run flask using the following file:

#!/usr/bin/env python
from datetime import datetime
         


        

Answer 1

One of the possible reason why the Flask app run itself twice is a configuration of WEB_CONCURRENCY setting on Heroku. To set into one, you can write in console heroku config:set WEB_CONCURRENCY=1

Answer 2

I had the same issue, and I solved it by setting app.debug to False. Setting it to True was causing my __name__ == "__main__" to be called twice.

Answer 3

The Werkzeug reloader spawns a child process so that it can restart that process each time your code changes. Werkzeug is the library that supplies Flask with the development server when you call app.run().

See the restart_with_reloader() function code; your script is run again with subprocess.call().

If you set use_reloader to False you'll see the behaviour go away, but then you also lose the reloading functionality:

app.run(port=4004, debug=config.DEBUG, host='0.0.0.0', use_reloader=False)

You can disable the reloader when using the flask run command too:

FLASK_DEBUG=1 flask run --no-reload

You can look for the WERKZEUG_RUN_MAIN environment variable if you wanted to detect when you are in the reloading child process:

import os
if os.environ.get('WERKZEUG_RUN_MAIN') == 'true':
    print '################### Restarting @ {} ###################'.format(
        datetime.utcnow())

However, if you need to set up module globals, then you should instead use the @app.before_first_request decorator on a function and have that function set up such globals. It'll be called just once after every reload when the first request comes in:

@app.before_first_request
def before_first_request():
    print '########### Restarted, first request @ {} ############'.format(
        datetime.utcnow())

Do take into account that if you run this in a full-scale WSGI server that uses forking or new subprocesses to handle requests, that before_first_request handlers may be invoked for each new subprocess.

Answer 4

If you are using the modern flask run command, none of the options to app.run are used. To disable the reloader completely, pass --no-reload:

FLASK_DEBUG=1 flask run --no-reload

Also, __name__ == '__main__' will never be true because the app isn't executed directly. Use the same ideas from Martijn's answer, except without the __main__ block.

if os.environ.get('WERKZEUG_RUN_MAIN') != 'true':
    # do something only once, before the reloader

if os.environ.get('WERKZEUG_RUN_MAIN') == 'true':
    # do something each reload

Answer 5

I had the same issue. I solved it by modifying my main and inserting use_reloader=False into it. If any body is here looking for a workaround for this problem then below code will get you started, however you will the functionality of changes in code being detected automatically by and restarting the application will not work. You will have to manually stop and restart you application after each edit in code.

if __name__ == '__main__':
    app.run(debug=True, use_reloader=False)

Answer 6

From Flask 0.11, it's recommended to run your app with flask run rather than python application.py. Using the latter could result in running your code twice.

As stated here :

... from Flask 0.11 onwards the flask method is recommended. The reason for this is that due to how the reload mechanism works there are some bizarre side-effects (like executing certain code twice...)