Python私有继承?如何欺骗isinstance()?
这是Django核心代码的一小段:
class ForeignKey(RelatedField, Field):
...
def db_type(self, connection):
rel_field = self.rel.get_related_field()
if (isinstance(rel_field, AutoField) or
(not connection.features.related_fields_match_type and
isinstance(rel_field, (PositiveIntegerField,
PositiveSmallIntegerField)))):
return IntegerField().db_type(connection=connection)
return rel_field.db_type(connection=connection)
这段代码很糟糕,因为如果我定义了一个自定义字段,它是从AutoField
继承来的,那么我的db_type方法就会被忽略。
我想做的是隐藏我的类是AutoField
的一个实例的事实。在C++中,我会通过私有继承来做到这一点。
有没有办法让isinstance
返回False
,或者隐藏继承关系呢?
这是我自定义字段的代码:
class MyAutoField(models.AutoField):
def __init__(self, length, *args, **kwargs):
self.length = length
super(MyAutoField, self).__init__(*args, **kwargs)
def db_type(self, connection):
if connection.vendor == 'oracle':
return 'NUMBER(%s,0)' % (self.length)
if connection.vendor == 'postgresql':
if self.length <= 4:
return 'smallint'
if self.length <= 9:
return 'integer'
return 'bigint'
return super(MyAutoField, self).db_type(connection)
1 个回答
6
我有一种方法可以做到这一点,但它是通过“猴子补丁”来实现的。你不能使用 ABCMeta
,因为那需要你重写基类的元类。
你可以像下面这样“改变” isinstance。重要的是“补丁”只需导入一次,我只会在没有其他办法时才这样做。
patches.py
import django.models
import mymodel
import __builtin__
def _isinstance(instance, clz):
if clz is models.AutoField and isinstance_orig(instance, MyAutoField):
return False
return isinstance_orig(instance, clz)
__builtin__.isinstance_orig = __builtin__.isinstance
__builtin__.isinstance = _isinstance
然后是你的测试程序:
class MyAutoField(models.AutoField): pass
x = MyAutoField()
print(isinstance(x, models.AutoField))
print(isinstance(x, models.MyAutoField))
介绍抽象基类: PEP-3119。下面是一个抽象的例子……
class ABCMeta(type):
def __instancecheck__(cls, inst):
"""Implement isinstance(inst, cls)."""
return any(cls.__subclasscheck__(c)
for c in {type(inst), inst.__class__})
def __subclasscheck__(cls, sub):
"""Implement issubclass(sub, cls)."""
candidates = cls.__dict__.get("__subclass__", set()) | {cls}
return any(c in candidates for c in sub.mro())