Django如何允许可交换模型的继承？

Question

我使用swappable来制作一个可重用的应用程序（名为Meat），它提供了开发人员可以交换的模型。那个型号是其他型号的超级。在

from django.db.models import Model, CharField
from swapper import swappable_setting

class AbstractMeat(Model):
    class Meta:
        abstract = True
    name = CharField(max_length=16)

class Meat(AbstractMeat):
    class Meta:
        swappable = swappable_setting("cyber", "Meat")

class Pork(Meat):
    pass

class Fish(Meat):
    pass

为了测试这一点，我创建了real应用程序并设置了MEAT_MEAT_MODEL。在

^{pr2}$

运行runserver我得到这个错误：

meat.Fish.meat_ptr: (fields.E301) Field defines a relation with the model 'meat.Meat', which has been swapped out.
    HINT: Update the relation to point at 'settings.MEAT_MEAT_MODEL'.
meat.Pork.meat_ptr: (fields.E301) Field defines a relation with the model 'meat.Meat', which has been swapped out.
    HINT: Update the relation to point at 'settings.MEAT_MEAT_MODEL'.

这个错误出现在django1.9到1.11上，但是对于我来说，只有1.11是关键的。在

我尝试按照Multi-table inheritance中的指示重写meat_ptr，如下所示：

from swapper import get_model_name
from django.db.models import OneToOneField, CASCADE

class Pork(Meat):
    meat_ptr = OneToOneField(
        get_model_name("meat", "Meat"), CASCADE,
        parent_link=True)

但它给出了1.11和1.10（但不是1.9）的错误：

django.core.exceptions.FieldError: Auto-generated field 'meat_ptr' in class 'Pork' for parent_link to base class 'Meat' clashes with declared field of the same name.

总之，我该怎么做呢？在

Answer 1

如果你想覆盖这个字段，看起来你必须把它称为肉以外的东西。自动生成的字段称为该字段。在

https://docs.djangoproject.com/en/1.11/topics/db/models/#multi-table-inheritance