Why Does My Flask Development Server Appear to Execute Twice?

Flask Development Server Executing Twice: Unveiling the Reason

The Flask development server handles the restarting of your application whenever changes are made to your code. This functionality is provided by the Werkzeug library's restart_with_reloader() function.

This function spawns a child process, executing your script again using subprocess.call(). Consequently, both the original script and the child process print the restart message once each, resulting in the observed double print behavior.

Disable the Reloader

To eliminate this behavior, disable the reloader by setting use_reloader to False:

app.run(port=4004, debug=config.DEBUG, host='0.0.0.0', use_reloader=False)

Alternatively, you can use the flask run command with the --no-reload flag:

FLASK_DEBUG=1 flask run --no-reload

Detect the Reloader Process

Use the werkzeug.serving.is_running_from_reloader() function to detect the child process:

from werkzeug.serving import is_running_from_reloader

if is_running_from_reloader():
    print(f"################### Restarting @ {datetime.utcnow()} ###################")

Use @app.before_first_request Decorator

If you need to set module globals, consider using the @app.before_first_request decorator:

@app.before_first_request
def before_first_request():
    print(f"########### Restarted, first request @ {datetime.utcnow()} ############")

This decorator executes the function once after every reload, when the first request is made.

Note:

In full-scale WSGI servers using forking or subprocesses to handle requests, before_first_request handlers may be invoked for each subprocess.

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