SQL炼金术:一对一的声明关系

如果你想要一对一的关系,你也必须在你的关系定义中使用“useList=false”。

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

0.7的文档 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'))

或

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)

我认为如果这是一对一的关系,我们应该为外键添加一个唯一性约束,这样另一个父级就不能有其他父级子级了!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)

事实证明这很容易。在您的foo模型中:

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