Why can't I debug my flask app with VSCode - flask

Question

from flask import Flask, render_template, request, url_for, g
app = Flask(__main__)
# This is the entrypoint
#app.route('/')
def index():
# Render page
return 'Hello'
if __name__=='__main__':
app.run(debug=True)
When I try to run this in the VSCode debugger I get:
* Serving Flask app 'gui' (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
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
* Restarting with stat
No module named gui
I set a breakpoint on the line "app.run(debug=True)" and it does execute to that point but when I try to step over or into I get the error.
I tried changing the name of the file from gui.py to app.py and I got the same thing except it says:
* 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
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
* Restarting with stat
No module named app
What am I doing wrong/missing?

Answer 1

To debug flask with VSCode you need to set your setting.json and launch.json files correctly, here is a possible configuration:
settings.json
{
"python.pythonPath": "<your_python_venv_path>",
}
launch.json
{
"version": "0.2.0",
"configurations" : [
{
"name": "local",
"type": "python",
"stopOnEntry": false,
"request": "launch",
"program" : "${workspaceFolder}/<path_to_run.py>",
"console" : "integratedTerminal",
"justMyCode" : false,
"cwd": "${workspaceFolder}"
}
]
}
Also take a look here