小编mik*_*ike的帖子

我试图在sqlalchemy中创建一个自引用的多对多关系(这意味着Line可以有许多父行和许多子行),如下所示:

Base = declarative_base()

class Association(Base):
 __tablename__ = 'association'

 prev_id = Column(Integer, ForeignKey('line.id'), primary_key=True)                            
 next_id = Column(Integer, ForeignKey('line.id'), primary_key=True)


class Line(Base):
 __tablename__ = 'line'

 id = Column(Integer, primary_key = True)
 text = Column(Text)
 condition = Column(Text)
 action = Column(Text)

 next_lines = relationship(Association, backref="prev_lines")



class Root(Base):
 __tablename__ = 'root'

 name = Column(String, primary_key = True)
 start_line_id = Column(Integer, ForeignKey('line.id'))

 start_line = relationship('Line')

但是我收到以下错误:sqlalchemy.exc.ArgumentError:无法确定关系Line.next_lines上的父/子表之间的连接条件.指定'primaryjoin'表达式.如果存在'secondary',则还需要'secondaryjoin'.

你知道我怎么能解决这个问题吗？