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数据库，但是当我运行脚本时出现此错误，我有另一个使用类似定义的项目，它以前工作过，但现在它现在出现相同的错误。 我使用的是Python 2.7.8，SQLAlchemy模块的版本是0.9.8。 顺便说一下，一个项目使用了Flask-SQLAlchemy，效果很好。 我很困惑。 回溯信息如下：

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)

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'))
)

使用声明式语法定义表时，元数据是通过类声明从Base继承的，即

Base = declarative_base()

class ChatLog(Base)

但是，当使用旧的 Table 语法定义表时，必须显式指定元数据。

Answer 2

我有同样的错误，因为我将

Column

拼写为小写

c

。应该是

Column

。

Answer 3

我也遇到过类似的问题。准确地说，我遇到的错误是：

AttributeError: Neither 'ColumnClause' object nor 'Comparator' object has an attribute '_set_parent_with_dispatch'

经过一番挣扎，我发现这是一个拼写错误。我在下面的代码片段行 #6 中将 sa.Column 拼写为 sa.column。

from alembic import op
import sqlalchemy as sa

def something():
    op.add_column('table_name',
        sa.Column('column_name', sa.Text(), nullable=True) # Correct usage
    )

Answer 4

我以一种非常愚蠢的方式收到了同样的错误消息。我写道：

Table(name=name, metadata=metadata, *columns)

当我应该写：

Table(name, metadata, *columns)