AssertionError: Expected a `Response`, `HttpResponse` or `HttpStreamingResponse` to be returned from the view, but received a `<class 'NoneType'>`

Question

我有以下以用户为成员的房间对象。

class Room(Base):
    name = models.CharField(db_index=True, unique=True, max_length=255)
    members = models.ManyToManyField(User, blank=True)

我正在尝试查找只有两个特定成员的房间，

if Room.objects.filter(members__id=first.id).filter(members__id=second.id).exists():
                    rooms = Room.objects.filter(members__id=first.id).filter(members__id=second.id)
                    for room in rooms:
                        print(room.members.count)
                        if room.members.count == 2:
                            return Response({"Success": RoomSerializer(room).data}, status=status.HTTP_200_OK)

我知道存在一个只有两个成员的 Room 对象。但我最终得到了这个错误，

AssertionError: Expected a `Response`, `HttpResponse` or `HttpStreamingResponse` to be returned from the view, but received a `<class 'NoneType'>`

感谢任何帮助。

Answer 1

如异常所述，在视图结束时，总是 return 响应。在你当前的代码中，如果逻辑不匹配，那么它 returns None.So，像这样更新代码：

if Room.objects.filter(members__id=first.id).filter(members__id=second.id).exists():
                rooms = Room.objects.filter(members__id=first.id).filter(members__id=second.id)
                for room in rooms:
                    print(room.members.count)
                    if room.members.count == 2:
                        return Response({"Success": RoomSerializer(room).data}, status=status.HTTP_200_OK)
return Response({"Failed": True}, status=status.HTTP_400_BAD_REQUEST) # <-- Return a bad request maybe at the end if all logic fails

Answer 2

QuerySet count() 是一种方法 (type(room.members.count) returns <class 'method'>)，应该这样调用。只是改变

room.members.count

到

room.members.count()

它应该按预期工作。

AssertionError: Expected a `Response`, `HttpResponse` or `HttpStreamingResponse` to be returned from the view, but received a `<class 'NoneType'>`

AssertionError: Expected a `Response`, `HttpResponse` or `HttpStreamingResponse` to be returned from the view, but received a `<class 'NoneType'>`

python

django

django-models

django-orm