我正在尝试在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),代码如下:
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')
但我得到了以下错误: sqlalchemy.exc.ArgumentError: 无法在关系Line.next_lines上确定父/子表之间的连接条件。请指定“primaryjoin”表达式。如果存在“secondary”,则还需要“secondaryjoin”。
你知道我该如何解决这个问题吗?