sqlalchemy中复合外键的多重关系

Question

我正在尝试创建一个边缘类型类来链接两个节点。每个节点本身就是一个类，节点和边都应该是同一个图的一部分。我不断得到

Could not determine join condition between parent/child tables on relationship TypeiNode.edges - there are no foreign keys linking these tables.

即使定义了外键。我尝试在节点表edges 关系上定义外键，但得到了相同的结果。有人能告诉我这个例子有什么问题吗？

class Edge(Base):
    __tablename__ = 'edge'

    id = Column(Integer, primary_key=True)
    graph_id = Column(Integer, ForeignKey('graph.id'))
    graph= relationship('Graph', back_populates='graph_edges')
    type_1_node_id = Column(Integer, ForeignKey('type_1_node.id'))
    type_1_node = relationship('Type1Node',
                               back_populates='edges',
                               foreign_keys=[type_1_node_id, graph_id])
    type_2_node_id = Column(Integer, ForeignKey('type_2_node.id'))
    type_2_node = relationship('Type2Node',
                               back_populates='edges',
                               foreign_keys=[type_2_node_id, graph_id ])
    __table_args__ = (
        ForeignKeyConstraint(
            ['type_1_node_id', 'graph_id'],
            ['type_1_node.id','type_1_node.graph_id']),
        ForeignKeyConstraint(
            ['type_2_node_id', 'graph_id'],
            ['type_2_node.id','type_2_node.graph_id']),
    )    

class Type1Node(Base):
    __tablename__ = 'type_1_node'

    id = Column(Integer, primary_key=True)
    graph_id = Column(Integer, ForeignKey('graph.id'))
    graph = relationship('Graph', back_populates='graph_1_nodes')
    edges = relationship('Edge', back_populates='type_1_node')

class Type2Node(Base):
    __tablename__ = 'type_2_node'

    id = Column(Integer, primary_key=True)
    graph_id = Column(Integer, ForeignKey('graph.id'))
    graph = relationship('Graph', back_populates='graph_2_nodes')
    edges = relationship('Edge', back_populates='type_2_node')

Answer 1

如果您仔细阅读该错误，您会发现它并没有抱怨您添加了 foreign_keys 参数的关系，而是抱怨您没有抱怨的关系：

Could not ... on relationship **TypeiNode.edges** ...

我接受

我尝试在节点表边缘关系上定义外键，但得到了相同的结果。

表示您没有尝试同时定义两端。所以解决方法是在两端定义外键：

class Type1Node(Base):
    __tablename__ = 'type_1_node'

    id = Column(Integer, primary_key=True)
    graph_id = Column(Integer, ForeignKey('graph.id'))
    graph = relationship('Graph')
    edges = relationship('Edge', back_populates='type_1_node',
                         foreign_keys=[Edge.type_1_node_id, Edge.graph_id])