开发者

How to create one-to-one relationships with declarative

开发者 https://www.devze.com 2023-01-11 09:43 出处:网络
I have two tables, foo and bar, and I want foo.bar_id to link to bar. The catch is that this is a one-way one-to-one开发者_开发技巧 relationship. bar must not know anything about foo. For every foo, t

I have two tables, foo and bar, and I want foo.bar_id to link to bar. The catch is that this is a one-way one-to-one开发者_开发技巧 relationship. bar must not know anything about foo. For every foo, there will be one and only one bar.

Ideally, after selecting a foo, I could do something like this:

myfoo.bar.whatever = 5 

How to accomplish this?


The documentation explains this nicely:

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

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)


If you want a true one-to-one relationship, you also have to use the "uselist=False" in your relationship definition.

bar_id = Column(Integer, ForeignKey(Bar.id))
bar = relationship(Bar, uselist=False)


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:

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)


It turns out this is actually quite easy. In your Foo model:

bar_id = Column(Integer, ForeignKey(Bar.id))
bar = relationship(Bar)
0

精彩评论

暂无评论...
验证码 换一张
取 消