How to run a flask program when debug = True

Question

I am a beginner to flask and I have been trying to run a simple program. For some reasons I am using THONNY as I just could not get flask running in a virtualenv even after so many tries. In thonny it worked without a virtualenv

Coming at the point I wrote a simple code

from flask import Flask

app = Flask(__name__)

@app.route('/')

def index():
    return 'Flask webapp'

if __name__ == "__main__":
    app.run(debug = True)

When I did not pass debug = True the app worked and was started on a localhost but after adding debug it gave me the following error

 * Serving Flask app 'app' (lazy loading)
 * Environment: production
   WARNING: This is a development server. Do not use it in a production deployment.
   Use a production WSGI server instead.
 * Debug mode: on
 * Restarting with stat
C:\Users\Workstation\AppData\Local\Programs\Thonny\python.exe: No module named thonny.plugins.cpython.app

What could be a possible fix to it? Should I continue using thonny for flask?

thank you for your response, my file name is app.py. it seems that there was some problem with the IDE itself, it was running on sublime without any problem — Rishi N, Jul 15 '21 at 08:57

score 1 · Answer 1 · answered Jul 14 '21 at 08:21

Install Flask by running (pip3 if using Python 3 and using linux) pip install flask - in a terminal.

When you install flask, run your code in a terminal by typing (py if you are using windows) py filename.py

(python or python3 if you are using linux or mac) python filename.py

How to run a flask program when debug = True

1 Answers1