我如何在sqlalchemy中获得循环关系？

Question

例如，我有两个彼此之间具有n：1关系的实体：

class Boy(MyBaseModel):
    __tablename__ = 'boys'
    name = Column(String, primary_key=True)
    crush_id = Column(String, ForeignKey('girls.name'))
    crush = relationship('Girl', foreign_keys=crush_id)


class Girl(MyBaseModel):
    __tablename__ = 'girls'
    name = Column(String, primary_key=True)
    crush_id = Column(String, ForeignKey('boys.name'))
    crush = relationship('Boy', foreign_keys=crush_id)

（（无意区别LGBT +乡亲，我只需要这个例子）

直到创建表，一切正常。在丘比特来袭的情况下，实际上我们有一个共同的迷恋，我明白了：

>>> b = Boy(name='foo')
>>> g = Girl(name='bar')
>>> 
>>> b.crush = g
>>> g.crush = b
>>> 
>>> session.add(b)
>>> session.add(g)
>>> session.commit()
[...]
sqlalchemy.exc.CircularDependencyError: Circular dependency detected. (ProcessState(ManyToOneDP(Girl.crush), <Girl at 0x7f4fe719c4d0>, delete=False), ProcessState(ManyToOneDP(Boy.crush), <Boy at 0x7f4fe7759d90>, delete=False), SaveUpdateState(<Girl at 0x7f4fe719c4d0>), SaveUpdateState(<Boy at 0x7f4fe7759d90>))

我该怎么做才能使它正常工作？ （无需更改表格）

Answer 1

好，我知道了。问题不在于模型定义中，而在于我试图一次提交两个关系。

这有效：

>>> b = Boy(name='foo')
>>> g = Girl(name='bar')
>>> b.crush = g
>>> session.add(b)
>>> session.add(g)
>>> session.commit()
>>> 
>>> g.crush = b
>>> session.commit()
>>> 
>>> b.crush
Girl(name=bar, crush_id=foo)
>>> b.crush.crush
Boy(name=foo, crush_id=bar)
>>>