Django 中间件 - 如何使用函数类型中间件的 process_view() 挂钩？

Question

我正在使用标准实现来实现一个新的中间件：

def my_middleware(get_response):
    def middleware(request):
        return get_response(request)
    return middleware

我想要view_args。我可以更改为基于 class 的中间件并实现方法

process_view(request, view_func, view_args, view_kwargs)

有没有其他方法可以在我的中间件中获取这些 view_args、view_kwargs 而无需更改为基于 class 的中间件？

Answer 1

将process_view(...)设置为middleware(...)的属性

<b>def process_view(request, view_func, view_args, view_kwargs):
    """
    do something
    """</b>


def simple_middleware(get_response):
    # One-time configuration and initialization.

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

        response = get_response(request)

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

        return response

    <b>middleware.process_view = process_view</b>

    return middleware

Django 中间件 - 如何使用函数类型中间件的 process_view() 挂钩？

Django middleware - How to use process_view() hook with function type middleware?

django

django-middleware