["python","flask"]
How to run a flask application?

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?
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().