Question

[Solved] Why can’t I change the host and port that my Flask app runs on?

I want to change the host and port that my app runs on. I set host and port in app.run, but the flask run command still runs on the default 127.0.0.1:8000. How can I change the host and port that the flask command uses?

if __name__ == '__main__':
    app.run(host='0.0.0.0', port=3000)
set FLASK_APP=onlinegame
set FLASK_DEBUG=true
python -m flask run

Solution #1:

The flask command is separate from the flask.run method. It doesn’t see the app or its configuration. To change the host and port, pass them as options to the command.

flask run -h localhost -p 3000

Pass --help for the full list of options.

Setting the SERVER_NAME config will not affect the command either, as the command can’t see the app’s config.


Never expose the dev server to the outside (such as binding to 0.0.0.0). Use a production WSGI server such as uWSGI or Gunicorn.

gunicorn -w 2 -b 0.0.0.0:3000 myapp:app
Respondent: davidism

Solution #2:

from flask import Flask
app = Flask(__name__)

@app.route("/")
def hello():
    return "Hello World!"

if __name__ == '__main__':
    app.run(host="localhost", port=8000, debug=True)

Configure host and port like this in the script and run it with

python app.py
Respondent: Gökhan Gerdan

Solution #3:

You can also use the environment variable FLASK_RUN_PORT, for instance:

export FLASK_RUN_PORT=8000
flask run
 * Running on http://127.0.0.1:8000/

Source: The Flask docs.

Respondent: Akronix

Solution #4:

When you run the application server using the flask run command, the __name__ of the module is not "__main__". So the if block in your code is not executed — hence the server is not getting bound to 0.0.0.0, as you expect.

For using this command, you can bind a custom host using the --host flag.

flask run --host=0.0.0.0

Source

Respondent: TeknasVaruas

Solution #5:

You also can use it:

if __name__ == "__main__":
    app.run(host='127.0.0.1', port=5002)

and then in the console use it

set FLASK_ENV=development
python app.py
Respondent: ????????? ???????

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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