Python 如何在 Django REST Framework 中返回自定义 JSON
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How to return custom JSON in Django REST Framework
提问by Roberto
I am trying to return custom json with get_querysetbut always get 404 errorin response.
我正在尝试返回自定义 json，get_queryset但总是得到404 error响应。
class TestViewSet(viewsets.ModelViewSet):
"""
API endpoint that allows groups to be viewed or edited.
"""
queryset = Test.objects.all()
serializer_class = TestSerializer
def get_queryset(self):
if self.request.method == "GET":
content = {'user_count': '2'}
return HttpResponse(json.dumps(content), content_type='application/json')
If I delete everything starting from defI'll got correct response with standard json data. What I am doing wrong?
如果我删除所有内容，def我将得到标准 json 数据的正确响应。我做错了什么？
采纳答案by bakkal
If you don't need a ModelViewSet and just want custom JSON on a GET request
如果您不需要 ModelViewSet 并且只想在 GET 请求上使用自定义 JSON
You can also use an APIView, which doesn't require a model
您也可以使用APIView，它不需要模型
class MyOwnView(APIView):
def get(self, request):
return Response({'some': 'data'})
and
和
urlpatterns = [
url(r'^my-own-view/$', MyOwnView.as_view()),
]
With a ModelViewSet
使用 ModelViewSet
You've put the custom JSON into get_queryset, that's wrong. If you want to use a ModelViewSet, this by itself should be enough:
您已将自定义 JSON 放入 get_queryset，这是错误的。如果你想使用 a ModelViewSet，这本身就足够了：
class TestViewSet(viewsets.ModelViewSet):
queryset = Test.objects.all()
serializer_class = TestSerializer
This ModelViewSetcomes with default implementations for .list(), .retrieve(), .create(), .update(), and .destroy(). Which are available for you to override (customize) as needed
这种ModelViewSet带有默认实现.list()，.retrieve()，.create()，.update()，和.destroy()。您可以根据需要覆盖（自定义）哪些
Returning custom JSON from .retrieve()and/or .list()in ModelViewSet
从返回的定制JSON.retrieve()和/或 .list()在ModelViewSet
E.g. to override .retrieve()to return custom view when retrieving a single object. We can have a look at the default implementation which looks like this:
例如，.retrieve()在检索单个对象时覆盖以返回自定义视图。我们可以看看默认的实现，它看起来像这样：
def retrieve(self, request, *args, **kwargs):
instance = self.get_object()
serializer = self.get_serializer(instance)
return Response(serializer.data)
So as an example to return custom JSON:
因此，作为返回自定义 JSON 的示例：
class TestViewSet(viewsets.ModelViewSet):
queryset = Test.objects.all()
serializer_class = TestSerializer
def retrieve(self, request, *args, **kwargs):
return Response({'something': 'my custom JSON'})
def list(self, request, *args, **kwargs):
return Response({'something': 'my custom JSON'})
回答by HoangYell
There are 2 ways to custom the response in Class-based viewswith ModelViewSet
有两种方法可以使用ModelViewSet在基于类的视图中自定义响应
Solution 1:custom in views.py
解决方案1：在views.py中自定义
class StoryViewSet(viewsets.ModelViewSet):
permission_classes = (permissions.AllowAny,)
queryset = Story.objects.all()
serializer_class = StorySerializer
def retrieve(self, request, *args, **kwargs):
# ret = super(StoryViewSet, self).retrieve(request)
return Response({'key': 'single value'})
def list(self, request, *args, **kwargs):
# ret = super(StoryViewSet, self).list(request)
return Response({'key': 'list value'})
Solution 2:custom in serializers.py(I recommend this solution)
解决方案2：在serializers.py中自定义（我推荐这个解决方案）
class StorySerializer(serializers.ModelSerializer):
class Meta:
model = Story
fields = "__all__"
def to_representation(self, instance):
ret = super(StorySerializer, self).to_representation(instance)
# check the request is list view or detail view
is_list_view = isinstance(self.instance, list)
extra_ret = {'key': 'list value'} if is_list_view else {'key': 'single value'}
ret.update(extra_ret)
return ret
