Question

我有三个表：用户，基金和基金类型。每个基金都有一个基金类型，每个用户都有一个基金列表，每个用户也可以拥有他们创建的基金类型列表。

架构：

class Fund(Base):
    __tablename__ = 'fds_funds'
    fds_fund_id = Column(Integer, primary_key=True)
    fds_name = Column(String(128))
    fds_symbol = Column(String(5))
    fds_fdt_fund_type_id = Column(Integer, ForeignKey('fdt_fund_type_id'))
    fund_type = relationship('FundType', backref=backref('fds_funds', uselist=False))


class FundType(Base):
    __tablename__ = 'fdt_fund_types'
    fdt_fund_type_id = Column(Integer, primary_key=True)
    fdt_type_name = Column(String(128))
    fdt_usr_user_id = Column(Integer, ForeignKey('usr_user_id'), nullable=True)


user_funds = Table('usf_user_funds', Base.metadata,
    Column('usf_usr_user_id', Integer, ForeignKey('usr_users.usr_user_id')),
    Column('usf_fds_fund_id', Integer, ForeignKey('fds_funds.fds_fund_id'))
)

class User(Base):
    """
    Application's user model.
    """
    __tablename__ = 'usr_users'
    usr_user_id = Column(Integer, primary_key=True)
    usr_email = Column(Unicode(50))
    _usr_password = Column('password', Unicode(64))

    fund_types = relationship('FundType', foreign_keys='FundType.fdt_usr_user_id')
    funds = relationship('Fund', secondary=user_funds)

我正在关注文档here，并且似乎非常明确地说外键指定的第一个参数应该是列名，而不是表名，但是我在收到此错误时我运行initialize_DB脚本：

sqlalchemy.exc.NoReferencedTableError：与列'fdt_fund_types.fdt_usr_user_id'关联的外键找不到表 'usr_user_id'用于生成目标列的外键 '无'

我误读了文档吗？

Answer 1

我不确定是什么修复它，但我做了一些重新安排和搞乱参考。对于后代，这有效：

user_funds = Table('usf_user_funds', Base.metadata,
    Column('usf_usr_user_id', Integer, ForeignKey('usr_users.usr_user_id')),
    Column('usf_fds_fund_id', Integer, ForeignKey('fds_funds.fds_fund_id'))
)

class User(Base):
    """
    Application's user model.
    """
    __tablename__ = 'usr_users'
    usr_user_id = Column(Integer, primary_key=True)
    usr_email = Column(Unicode(50))
    _usr_password = Column('password', Unicode(64))

    fund_types = relationship('FundType', foreign_keys='FundType.fdt_usr_user_id')
    funds = relationship('Fund', secondary=user_funds)   


class FundType(Base):
    __tablename__ = 'fdt_fund_types'
    fdt_fund_type_id = Column(Integer, primary_key=True)
    fdt_type_name = Column(String(128))
    fdt_usr_user_id = Column(Integer, ForeignKey('usr_users.usr_user_id'), nullable=True)

class Fund(Base):
    __tablename__ = 'fds_funds'
    fds_fund_id = Column(Integer, primary_key=True)
    fds_name = Column(String(128))
    fds_symbol = Column(String(5))
    fds_fdt_fund_type_id = Column(Integer, ForeignKey('fdt_fund_types.fdt_fund_type_id'))
    fund_type = relationship('FundType', backref=backref('funds', uselist=False))

不理解SQLalchemy上的错误消息

1 个答案: