SQLAlchemy循环一对一关系 [英] SQLAlchemy circular one-to-one relationship
问题描述
我正在尝试与看起来如下的SQLAlchemy建立一种循环的一对一关系(不确定正确的术语是什么):
I am trying to make a circular one-to-one relationship (not sure what the correct term is) with SQLAlchemy that looks the following:
class Parent(Base):
__tablename__ = 'parents'
id = db.Column(Integer, primary_key=True)
child_id = db.Column(db.Integer,db.ForeignKey("children.id", use_alter=True))
child = db.relationship("Child",
uselist=False,
foreign_keys=[child_id],
post_update=True)
class Child(Base):
__tablename__ = 'children'
id = db.Column(db.Integer, primary_key=True)
parent_id = db.Column(db.Integer, db.ForeignKey("parents.id"))
user = db.relationship("Parent",
uselist=False,
foreign_keys=[parent_id])
一切正常,直到我尝试db.drop_all()
并得到sqlalchemy.sql.schema.ForeignKeyConstraint
name
为None
的错误.尝试建立这种圆形一对一关系时,我做错什么了吗?我真的很想能够只查询单列以获取另一列的ID,因此可以使用循环引用.
Everything works as expected until I try to db.drop_all()
and I get an error that the sqlalchemy.sql.schema.ForeignKeyConstraint
name
is None
. Am I doing something wrong when trying to make this circular one-to-one relationship? I would really like to be able to query just the single column to get the id of the other one, hence the circular reference.
推荐答案
来自 SQLAlchemy文档:
class Parent(Base):
__tablename__ = 'parent'
id = Column(Integer, primary_key=True)
child = relationship("Child", uselist=False, back_populates="parent")
class Child(Base):
__tablename__ = 'child'
id = Column(Integer, primary_key=True)
parent_id = Column(Integer, ForeignKey('parent.id'))
parent = relationship("Parent", back_populates="child")
然后您可以全天使用Parent.child或Child.parent
Then you can Parent.child or Child.parent all day long
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