django - 在 Django 中间件中编辑响应内容

Question

我有 Django 1.10 项目和以下用户定义的中间件

class RequestLogMiddleWare(object):
    def __init__(self, get_response):
        self.get_response = get_response

    def __call__(self, request):
        response = self.get_response(request)
        response.data['detail'] = 'I have been edited'
        return response

和 REST 端点 View :

def r_mobile_call_log(request):
    return Response({'success': True, 
                     'detail': 'Before having been edited'}, 
                      status=status.HTTP_200_OK)

所以我希望客户端的最终响应是:

{'success': 'True', 'detail': 'I have been edited'}

然而，我看到的是:

{'success': 'True', 'detail': 'Before having been edited'}

我在中间件的 中放置了一个断点调用 方法来确保函数真的被执行了，就可以了。 response.data["details"]只是不会改变它的值(value)。有谁知道这是什么原因？

Answer 1

Response已经在中间件阶段呈现，所以你不能只是改变 response.data ，您需要重新渲染它或直接更改渲染内容。

class RequestLogMiddleWare(object):
    def __init__(self, get_response):
        self.get_response = get_response

    def __call__(self, request):
        response = self.get_response(request)
        if isinstance(response, Response):
            response.data['detail'] = 'I have been edited'
            # you need to change private attribute `_is_render` 
            # to call render second time
            response._is_rendered = False 
            response.render()
        return response

第二种方法是直接更改内容，但在这种情况下，内置的 REST 框架浏览器 API 将不起作用，因为模板将无法正确呈现。

import json

class RequestLogMiddleWare(object):
    def __init__(self, get_response):
        self.get_response = get_response

    def __call__(self, request):
        response = self.get_response(request)
        if isinstance(response, Response):
            response.data['detail'] = 'I have been edited'
            response.content = json.dumps(response.data)
        return response

source code for render method