Question

我想制作下一个模型架构：

class Place(models.Model):
    """
    This is base class for all places like shops, trc, restaurants, etc.
    """
    name = models.CharField(_('Place name'), max_length=255, unique=True, null=True, blank=True)
    partner = models.ForeignKey('Partner', verbose_name=_('Partner'), related_name='places')
    ...
    # next line causes errors
    trc = models.ForeignKey('Trc', related_name='places', null=True, blank=True) 


class Trc(Place):  # Trc is equal to Mall - place which contains shops, cafes, cinemas 
    ...

class Shop(Place):
    ...

class Restaurant(Place):
    ...

在Place模型中添加'trc'字段将使查询更好和通用，但是当我尝试在db中创建这样的表时，我收到以下错误：

CommandError: One or more models did not validate:
partner.trc: Accessor for field 'place_ptr' clashes with field 'Place.trc'. Add a related_name argument to the definition for 'place_ptr'.
partner.trc: Reverse query name for field 'place_ptr' clashes with field 'Place.trc'. Add a related_name argument to the definition for 'place_ptr'.

玩相关名称没有帮助......我真的不想在商店和餐厅中移动'trc'字段。谢谢

Answer 1

改为

class Place(models.Model):
    name = models.CharField(_('Place name'), max_length=255, unique=True, null=True, blank=True)
    partner = models.ForeignKey('Partner', verbose_name=_('Partner'), related_name='places')
    ...
    # next line causes errors
    #trc = models.ForeignKey('Trc', related_name='places', null=True, blank=True) 


class Trc(Place):  # Trc is equal to Mall - place which contains shops, cafes, cinemas 
    place = models.ForeignKey(Place, related_name='trc', ...)

等

Answer 2

无需添加trc字段，因为它已经存在。查看django docs on multi-table inheritance中非常相似的示例：

如果您的Place也是Restaurant，，您可以从 Place 对象转到 { {1}} 通过使用模型名称的小写版本：

Restaurant

但是，如果上例中的p = Place.objects.get(id=12) # If p is a Restaurant object, this will give the child class: p.restaurant <Restaurant: ...>不是p（原来是Restaurant 直接创建为Place对象或者是其他对象的父对象（），提到p.restaurant would提出一个 Restaurant.DoesNotExist例外。

ForeignKey从基类模型到继承者错误

2 个答案: