Fix middleware doc example (#1796)

I'm starting to use the project with the newly release 3 version.


70084bcc4c/docs/middleware.rst (L115-L120)

It seems this example is not correct. 

1. missing import for SecurityMiddleware
2. `ConnexionMiddleware.default_middlewares` is a list of classes, not
class instances: the filtering never happen.
This commit is contained in:
Aurélien Joga
2023-11-08 23:17:30 +01:00
committed by GitHub
parent 70084bcc4c
commit 095ae897a5

View File

@@ -113,9 +113,10 @@ Security through an API Gateway in front of your application, you can do:
.. code-block:: python
from connexion import AsyncApp, ConnexionMiddleware
from connexion.middleware.security import SecurityMiddleware
middlewares = [middleware for middleware in ConnexionMiddleware.default_middlewares
if not isinstance(middleware, SecurityMiddleware)]
if middleware is not SecurityMiddleware]
app = AsyncApp(__name__, middlewares=middlewares)
@@ -132,9 +133,10 @@ Security through an API Gateway in front of your application, you can do:
.. code-block:: python
from connexion import FlaskApp, ConnexionMiddleware
from connexion.middleware.security import SecurityMiddleware
middlewares = [middleware for middleware in ConnexionMiddleware.default_middlewares
if not isinstance(middleware, SecurityMiddleware)]
if middleware is not SecurityMiddleware]
app = FlaskApp(__name__, middlewares=middlewares)
@@ -153,9 +155,10 @@ Security through an API Gateway in front of your application, you can do:
from asgi_framework import App
from connexion import ConnexionMiddleware
from connexion.middleware.security import SecurityMiddleware
middlewares = [middleware for middleware in ConnexionMiddleware.default_middlewares
if not isinstance(middleware, SecurityMiddleware)]
if middleware is not SecurityMiddleware]
app = App(__name__)
app = ConnexionMiddleware(app, middlewares=middlewares)