django - 从中间件传递额外的参数来查看

Question

我想在我的 POST(JSON) 请求中添加另一个字段。键值对必须是“请求 ID”。我得出的结论是，最好的方法是生成一个随机请求 ID 并使用中间件将其注入(inject)到请求对象中。我为此编写了一个自定义中间件。当我尝试到达终点时出现错误。

我已经尝试通过互联网进行搜索，但没有找到解决我的错误的方法。

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.
    req = json.loads(request.body)
    req_id = random.randint(0,1000000)
    req["req_id"]=req_id

    response = self.get_response(req)

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

    return response

我收到的错误是“dict”对象没有属性“is_ajax”。你能帮我解决这个问题吗？如果有更简单更好的方法来实现这个，请告诉我。

Answer 1

好的。我实现了我想做的事情。我的自定义中间件代码是:

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
    
    def process_view(self, request, view_func, *view_args, **view_kwargs):
        request.req_id = random.randint(0,1000000)

更新:当时我是菜鸟，现在看到评论我觉得很尴尬:P