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我试图创建一个继承自超类中声明的元类的类，我确实这样做了：

from sqlalchemy import Column, String, create_engine
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker


class Database(object):
    """
        Class that holds all data models classes and functions
    """
    def __init__(self, wenDbPath = "test.db"):
        self.engine = create_engine('sqlite:///' + "c:/"+wenDbPath, echo=False)
        self.Session = sessionmaker(bind=self.engine)
        **self.Base** = declarative_base() # This returns a metaclass.
        self.Base.metadata.create_all(self.engine)
        self.session = self.Session()

    class Element(**self.Base**):
        __tablename__ = 'elements'
        idNumber = Column(String(255), primary_key=True)
        smallDiscription = Column(String(50), nullable=False)
        longDiscription = Column(String())
        webLink = Column(String())

        def __init__(self, idNumber, smallDiscription, longDiscription, webLink):
            self.idNumber = idNumber
            self.longDiscription = longDiscription
            self.smallDiscription = smallDiscription
            self.webLink = webLink
        def __repr__(self):
            return "<Element ('%s : %s')>" % (self.idNumber, self.smallDiscription)

Python给了我以下信息

类元素（self.Base）：NameError：名称'self'未定义

我怎么能做这样的事情？

先感谢您。

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1 回答 1

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__init__在运行之前评估类主体。由于Base不依赖于__init__参数，您可以在类评估时对其进行评估：

class Database(object):
    ...
    Base = declarative_base() # This returns a metaclass.
    class Element(Base):
        ...

请注意，您使用Base的是超类，而不是元类；元类语法是__metaclass__ = Base或(metaclass=Base)取决于版本。请参阅Python 中的元类是什么？

于 2013-02-14T16:58:16.087 回答