Question

我正在使用我的django app建立模型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，其中包含Player模型的所有字段。

现在我有两件事要做。

我可以为BookmarkPlayer模型扩展Player类。

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

我可以将Player模型的所有字段定义为BookmarkPlayer模型。

     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式更好。如果还有另一种好办法，请与我分享。

更新了问题

Knbb创建基类的想法很有意思，但我遇到的问题是我的一个模型已经存在于数据库中。

我的实际模特：

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 = True

class Player(BasePlayer):
   class Meta:
       app_label = 'core'


class BookmarkPlayer(BasePlayer):
    class Meta:
        app_label = 'core'

在我运行django服务器后进行这些更改后，我收到的错误如下所示。

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：
最后，如果我们将ForeignKey或ManyToManyField上的related_name属性用于Abstract模型，我会得到答案。
这通常会导致抽象基类出现问题，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）。
若要解决此问题，当您在抽象基类（仅）中使用related_name时，名称的一部分应包含＆＃39;％（app_label）s＆＃39;和＆＃39;％（类）s＆＃39;。
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需要相同的数据但需要在不同的表格中，那么抽象基础模型是最佳选择：

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都会从BasePlayer模型继承其字段，但由于BasePlayer是抽象的，因此模型完全解耦。

另一方面，多表继承仍然会将字段保存在单个表中，但是为BookmarkPlayer添加一个额外的表，并向OneToOneField表添加隐式Player。

Answer 2

这里有一些关于模型继承的好信息： https://docs.djangoproject.com/en/1.7/topics/db/models/#model-inheritance

选项2将引入重复，这总是很糟糕。如果BookmarkPlayer没有任何新字段，只有不同的方法，我建议您使用链接中描述的“代理模型”选项，因为您不需要BookmarkPlayer拥有自己的表数据库。

如果它需要它自己的表，“多表不干扰”是你要去的方式，这将是你问题中的选项1

创建一个包含当前现有模型的所有字段的新模型

2 个答案: