Python sqlalchemy：与声明式的一对一关系
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sqlalchemy: one-to-one relationship with declarative
提问by carl
What is the best way to create a one-to-one relationship in SQLAlchemy using declarative?
使用声明性在 SQLAlchemy 中创建一对一关系的最佳方法是什么？
I have two tables, fooand bar, and I want foo.bar_idto link to bar. The catch is that this is a one-way one-to-one relationship. barmust not know anything about foo. For every foo, there will be one and only one bar.
我有两个表foo和bar，我想foo.bar_id链接到bar. 问题在于，这是一种单向的一对一关系。bar一定一无所知foo。对于每个 foo ，都会有一个并且只有一个bar。
Ideally, after selecting a foo, I could do something like this:
理想情况下，选择 foo 后，我可以执行以下操作：
myfoo.bar.whatever = 5
What's the best way to accomplish this using declarative?
使用声明性完成此任务的最佳方法是什么？
采纳答案by Brett Bim
If you want a true one-to-one relationship, you also have to use the "uselist=False" in your relationship definition.
如果您想要真正的一对一关系，您还必须在关系定义中使用“uselist=False”。
bar_id = Column(Integer, ForeignKey(Bar.id))
bar = relationship(Bar, uselist=False)
回答by carl
It turns out this is actually quite easy. In your Foo model:
事实证明，这实际上很容易。在您的 Foo 模型中：
bar_id = Column(Integer, ForeignKey(Bar.id))
bar = relationship(Bar)
回答by chadwick.boulay
The documentation explains this nicely:
文档很好地解释了这一点：
class Parent(Base):
__tablename__ = 'parent'
id = Column(Integer, primary_key=True)
child = relationship("Child", uselist=False, backref="parent")
class Child(Base):
__tablename__ = 'child'
id = Column(Integer, primary_key=True)
parent_id = Column(Integer, ForeignKey('parent.id'))
OR
或者
class Parent(Base):
__tablename__ = 'parent'
id = Column(Integer, primary_key=True)
child_id = Column(Integer, ForeignKey('child.id'))
child = relationship("Child", backref=backref("parent", uselist=False))
class Child(Base):
__tablename__ = 'child'
id = Column(Integer, primary_key=True)
回答by hamidfzm
I think if it is a truly one to one relationship we should add a uniqueness constraint to foreign key so another parent can not have other parent child!! Like this:
我认为如果它是真正的一对一关系，我们应该为外键添加唯一性约束，这样另一个父级就不能有另一个父级子级！！像这样：
class Parent(Base):
__tablename__ = 'parent'
id = Column(Integer, primary_key=True)
child_id = Column(Integer, ForeignKey('child.id'), unique=True)
child = relationship("Child", backref=backref("parent", uselist=False))
class Child(Base):
__tablename__ = 'child'
id = Column(Integer, primary_key=True)
