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Why Does My Flask Development Server Print Startup Messages Twice?

Barbara Streisand
Release: 2024-12-20 01:04:08
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Why Does My Flask Development Server Print Startup Messages Twice?

Flask Development Server Running Twice

When running a Flask application using the app.run() method, you may encounter a situation where the print statement indicating the server restart appears twice. This behavior can be attributed to the Werkzeug reloader.

Werkzeug, the underlying library powering Flask's development server, employs a child process mechanism to facilitate code changes and auto-restarting. When you run app.run(), the reloader spawns a child process that consistently monitors your code.

To illustrate this, let's dissect the restart_with_reloader() function in Werkzeug. The function calls subprocess.call() to execute your script again, leading to the spawning of a child process. Consequently, you observe the print statement twice.

To eliminate this duplication, consider disabling the reloader. You can achieve this by setting use_reloader to False in app.run():

app.run(port=4004, debug=config.DEBUG, host='0.0.0.0', use_reloader=False)
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Alternatively, you can disable the reloader when using the flask run command:

FLASK_DEBUG=1 flask run --no-reload
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Another option is to utilize the werkzeug.serving.is_running_from_reloader() function to determine if you're executing within the reloader's child process.

However, if you require module globals, opt for the @app.before_first_request decorator. This decorator lets you specify a function that will be invoked once after each reload, when the first request is received:

@app.before_first_request
def before_first_request():
    print(f"########### Restarted, first request @ {datetime.utcnow()} ############")
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Note that when running in a full-scale WSGI server employing forking or subprocesses for request handling, before_first_request handlers might be invoked for each newly created subprocess.

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