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Python 试图从中间件功能中找出如何使用decorator\u_Python_Django_Decorator_Middleware - Fatal编程技术网

Python 试图从中间件功能中找出如何使用decorator\u

Python 试图从中间件功能中找出如何使用decorator\u,python,django,decorator,middleware,Python,Django,Decorator,Middleware,正如标题所述,我想在这里使用“decorator\u from\u middleware”函数: 然而,我只是对如何正确使用它感到困惑。我有我的自定义中间件类和所有正常的中间件设置。在装饰师的帮助下,我将如何使用这个函数来使用我的中间件作为每个视图的基础 例如: 假设我有一些中间件类 class SimpleMiddleware: def __init__(self, get_response): self.get_response = get_response

正如标题所述,我想在这里使用“decorator\u from\u middleware”函数:

然而,我只是对如何正确使用它感到困惑。我有我的自定义中间件类和所有正常的中间件设置。在装饰师的帮助下,我将如何使用这个函数来使用我的中间件作为每个视图的基础

例如: 假设我有一些中间件类

class SimpleMiddleware:
    def __init__(self, get_response):
        self.get_response = get_response
        # One-time configuration and initialization.

    def __call__(self, request):
        # Code to be executed for each request before
        # the view (and later middleware) are called.

        response = self.get_response(request)

        # Code to be executed for each request/response after
        # the view is called.

        return response

如何使用中间件(中间件类)中的装饰器?并将其应用于特定视图?

注意:您的中间件可能无法工作,因为文档断言它需要与旧式中间件方法兼容

假设此代码是您的视图模块:

from django.http import HttpResponse
from django.utils.decorators import decorator_from_middleware
from myapp.middleware import SimpleMiddleware

simple_decorator = decorator_from_middleware(SimpleMiddleware)

@simple_decorator 
def some_view(request):
    return HttpResponse("Hello World")
若要将此应用于基于类的视图,则需要修饰分派方法。您甚至可以这样编写一个mixin:

class SimpleMiddlewareMixin:
    @simple_decorator 
    def dispatch(*args, **kwargs):
        return super().dispatch(*args, **kwargs)

class MyClassBasedView(SimpleMiddlewareMixin, ListView):
    ...

非常感谢你!我会试试的这不行!