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我需要創建鏈表的SQLAlchemy版本。它實際上比這更復雜,但它歸結爲:SQLAlchemy中的一對一自我關係
我需要一個一對一,自我指涉,雙向關係在類中。每個元素可能只有一個父母,或者根本沒有,最多隻有一個孩子。
我簡化我的課,所以它看起來是這樣的:
class Node(declarative_base()):
__tablename__ = 'nodes'
id = Column(Integer, Sequence('nodes_id_seq'), primary_key=True, autoincrement=True)
value = Column(Integer)
prev_node_id = Column(Integer, ForeignKey('nodes.id'))
next = relationship('Node', uselist=False, backref=backref('prev', uselist=False))
然而,當我嘗試創建一個,它拋出一個異常:
>>> Node()
Traceback (most recent call last):
File "<console>", line 1, in <module>
File "<string>", line 2, in __init__
File "sqlalchemy\orm\instrumentation.py", line 309, in _new_state_if_none
state = self._state_constructor(instance, self)
[...]
File "sqlalchemy\orm\properties.py", line 1418, in _generate_backref
self._add_reverse_property(self.back_populates)
File "sqlalchemy\orm\properties.py", line 871, in _add_reverse_property
% (other, self, self.direction))
ArgumentError: Node.next and back-reference Node.prev are both of the same direction <symbol 'ONETOMANY>. Did you mean to set remote_side on the many-to-one side ?
缺少什麼我在這裏?谷歌搜索讓我絕對無處...:/
當我的id表沒有在類Node中定義但是在超類中時,如何設置remote_side? – user1111652
@ user1111652你可以在類之後聲明類似Node.prev = relationship(..... remote_side = [Node.id] .....) –
或者你可以使用帶引號的表達式來加載(類似於「 Node.id「而不是Node.id) –