Python SQLAlchemy：AttributeError：'Column' 对象和'Comparator' 对象都没有属性'schema'

Question

我尝试在我的项目中创建一个新的数据库，但是当我运行脚本时出现了这个错误，我有另一个使用类似定义的项目，它以前工作过，但现在出现了同样的错误。我使用的是 Python 2.7.8，SQLAlchemy 模块的版本是 0.9.8。顺便说一句，一个项目使用了Flask-SQLAlchemy，效果很好。我很迷惑。追溯信息如下：

 Traceback (most recent call last):
  File "D:/Projects/OO-IM/db_create.py", line 4, in <module>
    from models import Base
  File "D:\Projects\OO-IM\models.py", line 15, in <module>
    Column('followed_id', Integer(), ForeignKey('user.id'))
  File "C:\Python27\lib\site-packages\sqlalchemy\sql\schema.py", line 369, in __new__
    schema = metadata.schema
  File "C:\Python27\lib\site-packages\sqlalchemy\sql\elements.py", line 662, in __getattr__
    key)
AttributeError: Neither 'Column' object nor 'Comparator' object has an attribute 'schema'

from sqlalchemy import (create_engine, Column,
  String, Integer, Text,
  DateTime, Boolean,
  ForeignKey, Table)

from sqlalchemy.orm import sessionmaker, relationship, backref
from sqlalchemy.ext.declarative import declarative_base

SQLALCHEMY_DATABASE_URI = "mysql://root:mysqladmin@localhost:3306/oo_im?charset=utf8"

Base = declarative_base()

# TODO:AttributeError: Neither 'Column' object nor 'Comparator' object has an attribute 'schema'
friendships = Table('friendships',
                    Column('follower_id', Integer(), ForeignKey('user.id')),
                    Column('followed_id', Integer(), ForeignKey('user.id'))
)

class User(Base):
    __tablename__ = 'user'
    id = Column(Integer(), primary_key=True)
    account = Column(String(32), unique=True, nullable=False)
    password = Column(String(32), nullable=False)
    followed = relationship("User",
                            secondary=friendships,
                            primaryjoin=(friendships.c.follower_id == id),
                            secondaryjoin=(friendships.c.followed_id == id),
                            backref=backref("followers", lazy="dynamic"),
                            lazy="dynamic")

    def __init__(self, account, password, followed=None):
        self.account = account
        self.password = password

        if followed:
            for user in followed:
                self.follow(user)

    def follow(self, user):
        if not self.is_following(user):
            self.followed.append(user)
            return self

    def unfollow(self, user):
        if self.is_following(user):
            self.followed.remove(user)
            return self

    def is_following(self, user):
        return self.followed.filter(friendships.c.followed_id == user.id).count() > 0

class ChatLog(Base):
    __tablename__ = 'chatlog'
    id = Column(Integer(), primary_key=True)
    sender_id = Column(Integer(), ForeignKey('user.id'), nullable=False)
    receiver_id = Column(Integer(), ForeignKey('user.id'), nullable=False)
    send_time = Column(DateTime(), nullable=False)
    received = Column(Boolean(), default=False)
    content = Column(Text(), nullable=False)

engine = create_engine(SQLALCHEMY_DATABASE_URI, convert_unicode=True)
DBSession = sessionmaker(bind=engine)

原文由 elonzh 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

表定义应该是：

 friendships = Table('friendships',
                    Base.metadata,
                    Column('follower_id', Integer(), ForeignKey('user.id')),
                    Column('followed_id', Integer(), ForeignKey('user.id'))
)

当使用声明式语法定义表时，元数据通过类声明继承自Base，即

Base = declarative_base()

class ChatLog(Base)

但是，当使用旧的 Table 语法定义表时，必须明确指定元数据。

原文由 Haleemur Ali 发布，翻译遵循 CC BY-SA 3.0 许可协议

Answer 2

我有同样的错误，因为我用小写拼写了 Column c 。它应该是 Column 。

原文由 Hatshepsut 发布，翻译遵循 CC BY-SA 3.0 许可协议