Falcon recipe: Log an error whenever an error status code is returned

recipe.py

import logging

import falcon
import falcon.asgi
import falcon.util


class HelloResource:
    async def on_get(self):
        pass


class AuthMiddleware:
    async def process_request(self, req, resp):
        raise falcon.HTTPForbidden()

    async def process_response(self, req, resp, resource, req_succeeded):
        status_code = falcon.util.http_status_to_code(resp.status)
        if status_code >= 400:
            # NOTE: Could also report to an issue tracker such as Sentry.
            logging.error(f'Responding with an error status: {status_code}')


def create_app():
    app = falcon.asgi.App()
    app.add_middleware(AuthMiddleware())
    app.add_route('/', HelloResource())

    return app


app = create_app()