Question

我正在使用django rest框架。这是我的代码：

urls.py：

urlpatterns = [
    url(r'^users/show', UserShow.as_view()),
]

view.py：

class UserShow(ListAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer

    def get_queryset(self):
        queryset = User.objects.all()
        username = self.request.query_params.get('username', None)
        user_id = self.request.query_params.get('user_id', None)
        if username is not None:
            queryset = queryset.filter(username=username)
        if user_id is not None:
            queryset = queryset.filter(pk=user_id)
        return queryset

我想从url获取这样的值： /users/show?user_id=1或/users/show?username=mike。

user_id或username必须是必需参数。如何在基于类的视图中控制它？

如果我使用错误的参数名称/users/show?user111name=mike或简单/users/show发送请求，我的代码可以使用queryset = User.objects.all()来回复我并列出所有用户。我不需要那个。如果必需的参数是None响应404，我需要。

我可以通过基于功能的视图：

获得所需的结果

@api_view(['GET'])
def users(request):
    if request.method == 'GET':
        queryset = User.objects.all()
        username = request.GET.get('username', None)
        user_id = request.GET.get('user_id', None)

        if username is not None:
            queryset = queryset.filter(username=username)
        elif user_id is not None:
            queryset = queryset.filter(pk=user_id)
        else:
            return Response({"status": "required field not found."},
                            status=status.HTTP_404_NOT_FOUND)

        if not queryset.exists():
            return Response({"status": "not found."},
                            status=status.HTTP_404_NOT_FOUND)

        serializer = UserSerializer(queryset, many=True)
        return Response(serializer.data)

但是如何使用基于泛型类的视图进行操作？

Answer 1

class UserIdRetrieve(RetrieveAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer

class UserUsernameRetrieve(UserIdRetrieve):
     lookup_field = 'username'

并在网址中：

urlpatterns = [
    url(r'^users/(?P<pk>\d+)/', UserIdRetrieve.as_view()), 
    url(r'^users/by-username/(?P<username>\w+)/', UserUsernameRetrieve.as_view())
]

如果您的网址结构是必须的，小的更改为：

class UserIdRetrieve(RetrieveAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer

    def get_object(self):
        queryset = self.filter_queryset(self.get_queryset())

        if 'username' in self.request.query_params:
            filter_kwargs = {'username': self.request.query_params['username']}
        elif 'user_id' in self.request.query_params:
             filter_kwargs = {'id': self.request.query_params['user_id']}
        else:
            raise Http404('Missing required parameters')

        obj = get_object_or_404(queryset, **filter_kwargs)

        # May raise a permission denied
        self.check_object_permissions(self.request, obj)

        return obj

并在网址中：

urlpatterns = [
    url(r'^users/show', UserRetrieve.as_view())
]

Answer 2

class UserShow(ListAPIView):

    queryset = User.objects.all()
    serializer_class = UserSerializer

    def filter_queryset(self, queryset):
        username = self.request.query_params.get('username', None)
        user_id = self.request.query_params.get('user_id', None)

        if username is not None:
            queryset = queryset.filter(username=username)
        if user_id is not None:
            queryset = queryset.filter(pk=user_id)
        return queryset

    def list(self,request,*args,**kwargs):
        username = self.request.query_params.get('username', None)
        user_id = self.request.query_params.get('user_id', None)
        if not (username or user_id):
            return Response({"status": "Required field not found."},
                                        status=status.HTTP_404_NOT_FOUND)
        return super(UserShow, self).list(request,*args,**kwargs)

Django Rest需要URL中的参数

2 个答案: