Question

我有以下将用户作为成员的房间对象。

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

我正试图找到只有两个特定成员的会议室，

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Room对象，该对象只有两个成员。但是我最终遇到了这个错误，

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

任何帮助表示赞赏。

Answer 1

正如异常所说，在视图末尾，总是返回一个响应。在您当前的代码中，如果逻辑不匹配，则返回None。因此，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)返回<class 'method'>），应这样调用。只是改变

room.members.count

到

room.members.count()

它应该可以正常工作。

AssertionError：期望从视图返回“ Response”，“ HttpResponse”或“ HttpStreamingResponse”，但是收到了“ <class'nonetype'=“”>`

2 个答案: