【发布时间】:2010-10-27 21:41:15
【问题描述】:
我正在尝试这样做:
class Foo(Base):
id = Column(Integer, primary_key=True)
class Bar(Foo):
id = Column(Integer, primary_key=True)
class FooBarAssociation(Base):
foo_id = Column(Integer, ForeignKey('foo_table.id'))
bar_id = Column(Integer, ForeignKey('bar_table.id'))
foo = relationship(Foo, backref=...)
bar = relationship(Bar, backref=...)
...但我收到这样的错误:
Could not determine join condition between parent/child tables on relationship FooBarAssociation.foo. Specify a 'primaryjoin' expression. If this is a many-to-many relationship, 'secondaryjoin' is needed as well.
我尝试在关系声明中指定foreign_keys 和primary_join-s,但都是徒劳的。帮助? Foo 继承的 Bar 是不是惹到我了?
谢谢!
【问题讨论】:
标签: sqlalchemy