指定表间多个外键的默认连接路径

Question

代码

class User(Base):
    __tablename__ = 'user'
    id = Column(Integer, primary_key=True)
    name = Column(String)


class Asset(Base):
    __tablename__ = 'asset'
    id = Column(Integer, primary_key=True)
    user_id = Column(Integer, ForeignKey('user.id'))
    user = relationship(User, foreign_keys=[user_id])

    created_by = Column(Integer, ForeignKey(f"user.id"))
    updated_by = Column(Integer, ForeignKey(f"user.id"))


items = db.execute(select(User).join(Asset)).all()

抛出错误：

sqlalchemy.exc.AmbiguousForeignKeysError: Can't determine join between 'user' and 'asset'; tables have more than one foreign key constraint relationship between them. Please specify the 'onclause' of this join explicitly.

解决此问题的明显方法是在语句中指定要使用哪个外键：

items = db.execute(select(User).join(Asset.user)).all()

但是有没有办法通过改变模型来解决这个问题？

我想要类似的东西：

    created_by = Column(Integer, ForeignKey(f"user.id", do_not_use_for_joins=True))
    updated_by = Column(Integer, ForeignKey(f"user.id", do_not_use_for_joins=True))

或者至少：

    user = relationship(User, foreign_keys=[user_id], use_as_default_when_no_key_specified=True)

因为字段

created_by

和

updated_by

被添加到许多表中并且已经编写了很多代码。

这可能是一个意识形态问题。在模型中不将

created_by

和

updated_by

标记为外键（而是将它们保留在数据库中）是否正常？

Answer 1

我已经找到解决办法了

Modifier = aliased(User)


class Asset(Base):
    __tablename__ = 'asset'
    id = Column(Integer, primary_key=True)
    user_id = Column(Integer, ForeignKey('user.id'))
    user = relationship(User, foreign_keys=[user_id])

    created_by = Column(Integer, ForeignKey('Modifier.id'))
    updated_by = Column(Integer, ForeignKey('Modifier.id'))

    creator = relationship(Modifier, foreign_keys=[created_by], primaryjoin=created_by == Modifier.id)
    updater = relationship(Modifier, foreign_keys=[updated_by], primaryjoin=updated_by == Modifier.id)