Question

我有以下型号，用户

class User(AbstractBaseUser, PermissionsMixin, Base):
    email = models.EmailField(db_index=True, unique=True, max_length=255)
    mobile = PhoneNumberField(null=True)
    username = models.CharField(db_index=True, null=False, unique=True, max_length=255)

基类

class Room(Base):
    name = models.CharField(db_index=True, unique=True, max_length=255)
    status = models.CharField(default=RoomStatus.ACTIVE, max_length=256, null=True)
    members = models.ManyToManyField(User)
    last_activity = models.DateTimeField(default=timezone.now)

和两个孩子

class LeagueRoom(Room):
    league = models.ForeignKey(League, on_delete=models.CASCADE, null=True)
    location = models.ForeignKey(Location, on_delete=models.CASCADE, null=True)

class ClubRoom(Room):
    club = models.ForeignKey(Club, on_delete=models.CASCADE, null=True)
    location = models.ForeignKey(Location, on_delete=models.CASCADE, null=True)

我想为用户访问leagueroom_set。我不确定该怎么做。该用户当前具有属性room_set，但是我希望访问特定的room_set。有人可以帮我吗？

Answer 1

您可以从另一端开始：

league_rooms = LeagueRoom.objects.filter(members=my_user)

Answer 2

我认为您希望Room成为抽象的基类。然后，您尝试执行的操作将起作用。 https://docs.djangoproject.com/en/2.1/topics/db/models/#model-inheritance

class Room(models.Model):
    name = models.CharField(db_index=True, unique=True, max_length=255)
    # other fields
    …

    class Meta:
        abstract = True

如果Room不能是抽象类，则必须将Member字段从Room移到每个子类中才能使用内置关系。

class LeagueRoom(Room):
    league = models.ForeignKey(League, on_delete=models.CASCADE, null=True)
    location = models.ForeignKey(Location, on_delete=models.CASCADE, null=True)
    members = models.ManyToManyField(User)

# same with ClubRoom

我想到的第三种方法是在每个子类上设置一个type属性，以便可以将room_set过滤为所需的房间类型。
```
class LeagueRoom(Room):
   …
   type = models.CharField(max_length=10, default='league')
   …
```
然后在其他地方：
```
room_set.filter(type='league')
```

在Django中访问超类方法

2 个答案: