Question

我在尝试自我引用时遇到错误

class Sku(Base, ReprDescrIdMixin):
    __tablename__ = 'SC84'
    id = Column("ID", String, primary_key=True)
    parent_code = Column("PARENTID",String, ForeignKey('sku.id'))
    parent = relationship('Sku', foreign_keys='Sku.parent_code')


sqlalchemy.exc.NoForeignKeysError: Could not determine join condition between parent/child tables on relationship Sku.parent - there are no foreign keys linking these tables.  Ensure that referencing columns are associated with a ForeignKey or ForeignKeyConstraint, or specify a 'primaryjoin' expression.

为什么呢。这是工作http://docs.sqlalchemy.org/en/latest/orm/self_referential.html

Answer 1

试试这个

class Sku(Base, ReprDescrIdMixin):
    __tablename__ = 'SC84'
    id = Column("ID", String, primary_key=True)
    parent_code = Column("PARENTID",String, ForeignKey(id))
    parent = relationship('Sku', foreign_keys=parent_code)

使用ForeignKey(...)时，您应该传递参考密钥，有两种方法可以做到这一点

通过传递一个字符串，其中包含表格名称和与. ForeignKey("table_name.key")连接的键，如ForeignKey(key)
通过传递像key

如果ForeignKey(key)已经定义，那么您只能使用方法2，换句话说，如果key = Column(...)位于d3.queue之后

在我的示例中，我使用方法2，我更喜欢这种方式，以便我的IDE分析代码。

请参阅此link

Answer 2

自引用应该是表名（__tablename__）而不是对象名，以下是一对多的关系：

class Sku(Base, ReprDescrIdMixin):
    __tablename__ = 'SC84'
    id = Column("ID", String, primary_key=True)
    parent_code = Column("PARENTID",String, ForeignKey('SC84.id'))
    # ----------------------------------------------->  ^^^^ __tablename__ here
    children = relationship('Sku')

对于MANY-TO-ONE关系：

class Sku(Base, ReprDescrIdMixin):
    __tablename__ = 'SC84'
    id = Column("ID", String, primary_key=True)
    parent_code = Column("PARENTID",String, ForeignKey('SC84.id'))
    parent = relationship('Sku', remote_side=['id'])

Answer 3

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错误在哪里 - 邻接清单？

3 个答案: