如何为从 sqlalchemy 声明性基础扩展的类定义元类答案

【问题标题】：How to define metaclass for a class that extends from sqlalchemy declarative base如何为从 sqlalchemy 声明性基础扩展的类定义元类
【发布时间】：2013-06-30 14:31:31
【问题描述】：

我使用：Python 2.6 和 sqlalchemy 0.6.1

这就是我想要做的：

from sqlalchemy.types import (
    Integer,
    String,
    Boolean
)
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

class SampleMeta(type):
    def __new__(cls, name, bases, attrs):
        attrs.update({   'id': Column('Id', Integer, primary_key=True),
                    'name': Column('Name', String),
                    'description': Column('Description', String),
                    'is_active': Column('IsActive', Boolean)
                })
        return super(SampleMeta, cls).__new__(cls, name, bases, attrs)

class Sample(Base):
    __tablename__ = 'Sample'
    __table_args__ = {'useexisting': True}
    __metaclass__ = SampleMeta

    def __init__(self, id, name, description, is_active):
        self.id = id
        self.name = name
        self.description = description
        self.is_active = is_active

    def __repr__(self):
        return "<(%d, '%s', '%s', %r)>" % (self.id, self.name, self.description, self.isactive)

我得到的错误是：

TypeError: Error when calling the metaclass bases
    metaclass conflict: the metaclass of a derived class must be a (non-strict) subclass of the metaclasses of all its bases

现在，如果我使用上面的方法做同样的事情

class Sample(object)

而不是

class Sample(Base)

它工作得很好。

我需要动态更新类的属性。所以，我将使用动态属性和列名。我需要上面的代码才能到达那里。

请帮忙

【问题讨论】：

标签： python dynamic attributes sqlalchemy metaclass

【解决方案1】：

要使用元类，在这种情况下，它必须从声明式DeclaredMeta 派生。但是这个用例不需要元类。使用mixins:

from sqlalchemy.types import (
    Integer,
    String,
    Boolean
)
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import Column

Base = declarative_base()

class SampleMixin(object):
    id = Column("Id", Integer, primary_key=True)
    name = Column("Name", String)
    description = Column("Description", String)
    is_active = Column("IsActive", Boolean)

class Sample(SampleMixin, Base):
    __tablename__ = 'Sample'
    __table_args__ = {'useexisting': True}

    def __init__(self, id, name, description, is_active):
        self.id = id
        self.name = name
        self.description = description
        self.is_active = is_active

    def __repr__(self):
        return "<(%d, '%s', '%s', %r)>" % (self.id, self.name,
                                self.description, self.isactive)

from sqlalchemy import select
print select([Sample.id, Sample.name])

【讨论】：

【解决方案2】：

如果 zzzeek 提出的多重继承不能解决您的问题，请尝试从 DeclarativeMeta 继承您的 MetaClass：

from sqlalchemy.ext.declarative import declarative_base, DeclarativeMeta

Base = declarative_base()

class SampleMeta(DeclarativeMeta):
    #...

class Sample(Base):
    __metaclass__ = SampleMeta
    #...

【讨论】：