What is the best way to programmatically know what module to set FLASK_APP to?

[brettcannon]

For the Python extension in VS Code, we want to provide automatic debug configurations for Flask users when we can easily do so. That means (we think) needing to know what module to set FLASK_APP to. Is there a best practice for determining this or something we are missing that we can use to determine this? Or is there something else we can rely on to do the right thing where we can control how Python is executed?

[ThiefMaster]

Some ideas that come to my mind:

• Look for a module that contains app = Flask(...) (ie something named app with isinstance(app, Flask))
• Look for a module that contains a create_app() or init_app() function that returns a Flask instance
• Look for a .env or .flaskenv that sets FLASK_APP and use that

[brettcannon]

Some ideas that come to my mind

Thanks for the ideas!

• (ie something named app with isinstance(app, Flask))

We would like to avoid executing Python code since the debugger is launched via TypeScript and a created shell command. But we could run a regex over some files to look for app\s*=\s*Flask\(.

[ThiefMaster]

I was more thinking something in the typing/mypy direction to figure out if something is a Flask instance (or instance of a Flask subclass) ;)

[davidism]

Flask itself has some detection for when FLASK_APP isn't set that will probably serve you well enough.

1. Look for a file called wsgi.py or app.py in the current directory.
2. Within that, look for a variable called app or application, if it's a Flask object return it.
3. Look for any variable that is an instance of Flask and return that.
4. Look for a function called create_app or make_app. Call it without arguments, if it returns a Flask object return it.

I'd suggest extending this to looking in __init__.py and app.py files within the packages in the current project, something VS Code should be able to figure out which Flask itself can't. And since you're not actually executing code, look for var = Flask( like you suggested, and assume factory functions return correctly.

If you can't detect something automatically using the rules above, or even if you can, I'd suggest looking at PyCharm Professional's Flask server run configuration. It can detect what to run in many cases (probably the rules above), and then allows customizing them further to support the full range of what FLASK_APP and flask run can do.

[brettcannon]

What we are thinking of doing is searching for ^app(?:lication)?\s*=\s*=(?:flask\.)Flask\( or def\s+(?:create|make)_app\( in wsgi.py, app.py, */__init__.py, */app.py, src/*/__init__.py, and src/*/app.py.

[davidism]

@brettcannon not sure if this is handled by VS Code, but it's related to running the development server. I'm deprecating the FLASK_ENV=development environment variable and going back to using FLASK_DEBUG=1 (or the flask --debug run option as well now) to control debug mode. If you're generating a run configuration, you might want to enable debug mode by default.

If you're running with the VS Code debugger, you might want to disable the Flask debugger and reloader as well, using the flask --debug run --no-debugger --no-reload options. All options can also be controlled with environment variables, like FLASK_RUN_DEBUGGER=0 and FLASK_RUN_RELOAD=0.

[paulacamargo25]

Thanks for you reply @davidism, I was going through the arguments we have in the python extension debugger and wanted to ask why in the previous comment you suggested disabling the Flask debugger and reloader. Since on the user side it is better if these are activated. So I wanted to confirm if there is a known bug for leaving these options active, so I can leave them active by default.

Thanks

[davidism]

If you're using the IDE's debugger, you don't also need the built-in debugger UI. If you're using breakpoints, you probably don't want the server to restart in the middle of a debugging session. Both can remain enabled, they just also make sense disabled in this case.