[Solved]-Editing response content in Django middleware

Response is already rendered in the middleware stage so you can’t just change response.data, you need to rerender it or change rendered content directly.

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

The second approach is to change content directly, but in that case builtin REST Framework browser API will not work because template will not render properly.

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

I don’t suggest editing REST-Framework’s Response with using middleware. I think you should overload REST-Framework’s default JSONRenderer.

Define a custom renderer:

from rest_framework.renderers import JSONRenderer


class CustomJsonRender(JSONRenderer):

    def render(self, data, accepted_media_type=None, renderer_context=None):

        if renderer_context:
            response = renderer_context['response']
            msg = "OK"
            code = response.status_code
            if isinstance(data, dict):
                msg = data.pop('msg', msg)
                code = data.pop('code', code)
                data = data.pop('data', data)
            if code != 200 and data:
                msg = data.pop('detail', 'failed')
            response.status_code = 200
            res = {
                'code': code,
                'msg': msg,
                'data': data,
            }
            return super().render(res, accepted_media_type, renderer_context)
        else:
            return super().render(data, accepted_media_type, renderer_context)

Using it in you APiVIew or ViewSet:

class SimpleView(APIView):
    renderer_classes = (CustomJsonRender,)
    def get(self, request):
        # do something
        return Response({"id":"xx"})

class SimpleViewSet(ModelViewSet):
    renderer_classes = (CustomJsonRender,)
    queryset = SomeModel.objects.all()
    serializer_class = SomeSerializer

You can edit it golbal in settings as well:

REST_FRAMEWORK = {
    "DEFAULT_RENDERER_CLASSES": ("compent.renders.CustomJsonRender",)
}

Then your response has been changed.

I have a feeling that I found cleaner solution. Here’s how I rewrote the code:

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

    def __call__(self, request):
       response = self.get_response(request)
       return response

    def process_template_response(self, request, response):
       if hasattr(response, 'data'): 
          response.data['detail'] = 'bla-bla-bla'
       return response

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)

    res = str(response.content)

    res += "<h2>add some html!</h2>"

    response.content = bytes(res, encoding="UTF8")

    return response

response.content type is bytes, so you need first convert it to str,

add something to it, convert it to bytes

and finally assign it to the response.content.