I want to know the correct way to start a flask application. The docs show two different commands:
$ flask -a sample run
and
$ python3.4 sample.py
produce the same result and run the application correctly.
What is the difference between the two and which should be used to run a Flask application?
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Answer
The flask
command is a CLI for interacting with Flask apps. The docs describe how to use CLI commands and add custom commands. The flask run
command is the preferred way to start the development server.
Never use this command to deploy publicly, use a production WSGI server such as Gunicorn, uWSGI, Waitress, or mod_wsgi.
As of Flask 2.2, use the --app
option to point the command at your app. It can point to an import name or file name. It will automatically detect an app instance or an app factory called create_app
. Use the --debug
option to run in debug mode with the debugger and reloader.
$ flask --app sample --debug run
Prior to Flask 2.2, the FLASK_APP
and FLASK_ENV=development
environment variables were used instead. FLASK_APP
and FLASK_DEBUG=1
can still be used in place of the CLI options above.
$ export FLASK_APP=sample $ export FLASK_ENV=development $ flask run
On Windows CMD, use set
instead of export
.
> set FLASK_APP=sample
For PowerShell, use $env:
.
> $env:FLASK_APP = "sample"
The python sample.py
command runs a Python file and sets __name__ == "__main__"
. If the main block calls app.run()
, it will run the development server. If you use an app factory, you could also instantiate an app instance at this point.
if __name__ == "__main__": app = create_app() app.run(debug=True)
Both these commands ultimately start the Werkzeug development server, which as the name implies starts a simple HTTP server that should only be used during development. You should prefer using the flask run
command over the app.run()
.