python - URL 中的 Django Rest 所需参数

Question

我正在使用 Django 休息框架。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

我想像这样从 url 中获取值:/users/show?user_id=1 或 /users/show?username=mike。

user_id 或 username 必须是必需的参数。 如何在基于类的 View 中控制它？

如果我使用错误的参数名称发送请求 /users/show?user111name=mike 或简单的 /users/show 类(class)响应 View 我使用 queryset = User.objects.all() 并列出所有用户。我不需要那个。如果所需参数是 None 响应，我需要 404。

我可以通过基于函数的 View 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)

但是我如何使用基于通用类的 View 来做到这一点？

Answer 1

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

class UserUsernameRetrieve(UserIdRetrieve):
     lookup_field = 'username'

在 url 中:

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

如果你的 url 结构是必须的，对上面的小改动:

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

在 url 中:

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