问题
I need to create a class that uses a different base class depending on some condition. With some classes I get the infamous:
TypeError: metaclass conflict: the metaclass of a derived class must be a (non-strict) subclass of the metaclasses of all its bases
One example is sqlite3, here is a short example you can even use in the interpreter:
>>> import sqlite3
>>> x = type('x', (sqlite3,), {})
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: metaclass conflict: the metaclass of a derived class must be a (non-strict) subclass of the metaclasses of all its bases
回答1:
Instead of using the receipe as mentioned by jdi, you can directly use:
class M_C(M_A, M_B):
pass
class C(A, B):
__metaclass__ = M_C
回答2:
Your example using sqlite3 is invalid because it is a module and not a class. I have also encountered this issue.
Heres your problem: The base class has a metaclass that is not the same type as the subclass. That is why you get a TypeError.
I used a variation of this activestate snippet using noconflict.py. The snippet needs to be reworked as it is not python 3.x compatible. Regardless, it should give you a general idea.
Problem snippet
class M_A(type):
pass
class M_B(type):
pass
class A(object):
__metaclass__=M_A
class B(object):
__metaclass__=M_B
class C(A,B):
pass
#Traceback (most recent call last):
# File "<stdin>", line 1, in ?
#TypeError: metaclass conflict: the metaclass of a derived class must be a (non-strict) subclass #of the metaclasses of all its bases
Solution snippet
from noconflict import classmaker
class C(A,B):
__metaclass__=classmaker()
print C
#<class 'C'>
The code recipe properly resolves the metaclasses for you.
回答3:
To use the pattern described by @michael, but with both Python 2 and 3 compatibility (using the six library):
from six import with_metaclass
class M_C(M_A, M_B):
pass
class C(with_metaclass(M_C, A, B)):
# implement your class here
回答4:
This also happens when you try to inherit from a function and not a class.
Eg.
def function():
pass
class MyClass(function):
pass
回答5:
As far as I understood from the previous answers the only think we usually have to do manually is:
class M_A(type): pass
class M_B(type): pass
class A(metaclass=M_A): pass
class B(metaclass=M_B): pass
class M_C(M_A, M_B): pass
class C:(A, B, metaclass=M_C): pass
But we can automate the last two lines now by:
def metaclass_resolver(*classes):
metaclass = tuple(set(type(cls) for cls in classes))
metaclass = metaclass[0] if len(metaclass)==1 \
else type("_".join(mcls.__name__ for mcls in metaclass), metaclass, {}) # class M_C
return metaclass("_".join(cls.__name__ for cls in classes), classes, {}) # class C
class C(metaclass_resolver(A, B)): pass
Since we do not use any version-specific metaclass syntax this metaclass_resolver works with Python 2 as well as Python 3.
来源:https://stackoverflow.com/questions/11276037/resolving-metaclass-conflicts