Failing to start (due to port unavailable) cannot be directly determined by the current python api about flask HOT 4 CLOSED

Complete code reproducing the context, which would hopefully also become a skeleton (or inspiration) for a flaks test, after an api transparency into failure to use the specified server port has been very hopefully incorporated in flask:

from flask import Flask
from werkzeug.serving import run_simple
from multiprocessing import Process

app = Flask(__name__)

def start():
    print('starting a flask server')
    try:
        started = run_simple('127.0.0.1', 5000, app)
    except Exception as e:
        print(f'the flask web server crashed with an exception:\n{e}\n')
    else:
        # of course we will never get here unless the server shut down without throwing
        print(f'the flask web server finished with return value {started}')


if __name__ == '__main__':

    flask_process = Process(target=start, daemon=True)
    flask_process.start()

    start()

Note that, the messages about the port being unavailable come out on stderr, same as success messages such as the following type, and hence this leaves no deterministic way for the code starting the server to resolve the status of the server, other than parsing stderr:

127.0.0.1 - - [28/Apr/2024 13:31:24] "GET / HTTP/1.1" 200 -

from flask.

Not sure what you are trying to do, but keep in mind that this server is ONLY meant for development purposes, NOT for anything else. So if you want to run a flask app embedded in some other application for whatever reason, using run_simple is not what you should do.

from flask.

Makes sense. My scenario is borderline in that regard. Will I have that kind of api-laden way of knowing the status of the server when using a non Werkzeug WSGI component?

from flask.

Okay, I understand then that no underlying supported WSGI server component can be controlled (start and stop) robustly from Flask code then, and that that is not planned either.

from flask.