python - 创建一个新模型，其中包含当前现有模型的所有字段

Question

我的 django 应用程序有一个模型 Player。

class Player(models.Model):
    """ player model """
    name = models.CharField(max_length=100, null=True, blank=True)
    date_created = models.DateTimeField(auto_now_add=True)
    last_updated = models.DateTimeField(auto_now=True)
    hash = models.CharField(max_length=128, null=True, blank=True)
    bookmark_url = models.CharField(max_length=300, null=True, blank=True)

根据我的要求，我需要创建一个新模型 BookmarkPlayer，它将包含 Player 模型的所有字段。

现在我有两件事要做。

1. 我可以为 BookmarkPlayer 模型扩展 Player 类。

    class BookmarkPlayer(Player):        """ just a bookmark player"""        class Meta:            app_label = "core"

1. 我可以将 Player 模型的所有字段定义到 BookmarkPlayer 模型中。

     class BookmarkPlayer(models.Model):            """ bookmark player model """            name = models.CharField(max_length=100, null=True, blank=True)            date_created = models.DateTimeField(auto_now_add=True)            last_updated = models.DateTimeField(auto_now=True)            hash = models.CharField(max_length=128, null=True, blank=True)            bookmark_url = models.CharField(max_length=300, null=True, blank=True)

我只是想知道哪种方法更好，如果有其他好的方法，请与我分享。

更新问题

Knbb 创建基类的想法很有趣，但我的一个模型已存在于数据库中，但我遇到了问题。

我的实际模型:

class Address(models.Model):    address = models.TextField(null=True, blank=True)class Site(models.Model):    domain = models.CharField(max_length=200)class Player(models.Model):    # ... other fields    shipping_address = models.ForeignKey(Address, related_name='shipping')    billing_address = models.ForeignKey(Address, related_name='billing')    created_at = models.DateTimeField(auto_now_add=True)    updated_at = models.DateTimeField(auto_now_add=True)    site = models.ManyToManyField(Site, null=True, blank=True)    class Meta:       abstract = True

更改后的模型:

class Address(models.Model):    address = models.TextField(null=True, blank=True)class Site(models.Model):    domain = models.CharField(max_length=200)class BasePlayer(models.Model):    # .. other fields    shipping_address = models.ForeignKey(Address, related_name='shipping')    billing_address = models.ForeignKey(Address, related_name='billing')    created_at = models.DateTimeField(auto_now_add=True)    updated_at = models.DateTimeField(auto_now_add=True)    site = models.ManyToManyField(Site, null=True, blank=True)    class Meta:       abstract = Trueclass Player(BasePlayer):   class Meta:       app_label = 'core'class BookmarkPlayer(BasePlayer):    class Meta:        app_label = 'core'

进行这些更改后，当我运行我的 Django 服务器时，出现以下错误。

django.core.management.base.CommandError: One or more models did not validate:core.test1: Accessor for field 'shipping_address' clashes with related field 'Address.shipping'. Add a related_name argument to the definition for 'shipping_address'.core.test1: Reverse query name for field 'shipping_address' clashes with related field 'Address.shipping'. Add a related_name argument to the definition for 'shipping_address'.core.test1: Accessor for field 'billing_address' clashes with related field 'Address.billing'. Add a related_name argument to the definition for 'billing_address'.core.test1: Reverse query name for field 'billing_address' clashes with related field 'Address.billing'. Add a related_name argument to the definition for 'billing_address'.core.test2: Accessor for field 'shipping_address' clashes with related field 'Address.shipping'. Add a related_name argument to the definition for 'shipping_address'.core.test2: Reverse query name for field 'shipping_address' clashes with related field 'Address.shipping'. Add a related_name argument to the definition for 'shipping_address'.core.test2: Accessor for field 'billing_address' clashes with related field 'Address.billing'. Add a related_name argument to the definition for 'billing_address'.core.test2: Reverse query name for field 'billing_address' clashes with related field 'Address.billing'. Add a related_name argument to the definition for 'billing_address'

答案:
如果我们在抽象模型中使用 ForeignKey 或 ManyToManyField 上的 related_name 属性，我终于得到了答案。
这通常会在抽象基类中引起问题，因为此类中的字段包含在每个子类中，每次都具有完全相同的属性值(包括 related_name)。
要解决此问题，当您(仅)在抽象基类中使用 related_name 时，名称的一部分应包含“%(app_label)s”和“%(class)s”。
https://docs.djangoproject.com/en/dev/topics/db/models/#abstract-base-classes
现在我的 BasePlayer 模型是

class BasePlayer(models.Model):    # .. other fields    shipping_address = models.ForeignKey(Address, related_name='%(app_label)s_%(class)s_shipping')    billing_address = models.ForeignKey(Address, related_name='%(app_label)s_%(class)s_billing')    created_at = models.DateTimeField(auto_now_add=True)    updated_at = models.DateTimeField(auto_now_add=True)    site = models.ManyToManyField(Site, null=True, blank=True)    class Meta:       abstract = True

Answer 1

如果您的 BookmarkPlayer 需要相同的数据但在不同的表中，抽象基础模型是最好的方法:

class BasePlayer(models.Model):
    name = models.CharField(max_length=100, null=True, blank=True)
    date_created = models.DateTimeField(auto_now_add=True)
    last_updated = models.DateTimeField(auto_now=True)
    hash = models.CharField(max_length=128, null=True, blank=True)
    bookmark_url = models.CharField(max_length=300, null=True, blank=True)

    class Meta:
        abstract = True

class Player(BasePlayer):
    """ player model """
    pass

class BookmarkPlayer(BasePlayer):
    """ bookmark player model """
    pass

这样，Player 和 BookmarkPlayer 都从 BasePlayer 模型继承了它们的字段，但是因为 BasePlayer 是抽象，模型完全解耦。

另一方面，多表继承仍会将字段保存在单个表中，但会为 BookmarkPlayer 添加一个额外的表，并将隐式 OneToOneField 添加到 Player 表。