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When i'm running 'flask run' from the terminal window, I'm getting '* Serving Flask-SocketIO app "application"', however I don't get further provided with an URL!?

Anyone any thought on that what the problem might be?

ps: I followed precisely all the steps as outlined in the description: downloaded and unzipped project2, in a terminal window, navigated into my project2 directory, ran pip3 install -r requirements.txt, set the environment variable FLASK_APP to be application.py as recommended for MacBook with export., ran 'flask run'

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  • I had the same thing... I installed asynchronous service eventless and uninstalled gevent. Took care of it.
    – mobert
    Commented Aug 10, 2018 at 13:42

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Even if you are not getting url you can go this http://127.0.0.1:5000 your server is running and if you still not getting the results then add this code

if __name__ == '__main__':
socketio.run(app)

just after the socketio = SocketIO(app) line and execute flask run command

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