Question

我有一个Location类。位置可以具有默认的“账单到地址”，这也是位置。我正在使用的字段是CustomerLocation类中的bill_to_id和bill_to。为了完整性，我已将父类包括在内。如何将一个位置设置为另一个位置的账单？这种关系应该是一对一的（一个地点只有一个账单）。不需要反射。

TIA

class Location(DeclarativeBase,TimeUserMixin):
    __tablename__ = 'locations'

    location_id = Column(Integer,primary_key=True,autoincrement=True)
    location_code = Column(Unicode(10))
    name = Column(Unicode(100))
    address_one = Column(Unicode(100))
    address_two = Column(Unicode(100))
    address_three = Column(Unicode(100))
    city = Column(Unicode(100))
    state_id = Column(Integer,ForeignKey('states.state_id'))
    state_relate = relation('State')
    zip_code = Column(Unicode(100))
    phone = Column(Unicode(100))
    fax = Column(Unicode(100))
    country_id = Column(Integer,ForeignKey('countries.country_id'))
    country_relate = relation('Country')
    contact = Column(Unicode(100))
    location_type = Column('type',Unicode(50))

    __mapper_args__ = {'polymorphic_on':location_type}

class CustomerLocation(Location):
    __mapper_args__ = {'polymorphic_identity':'customer'}
    customer_id = Column(Integer,ForeignKey('customers.customer_id',
                                            use_alter=True,name='fk_customer_id'))
    customer = relation('Customer',
                        backref=backref('locations'),
                        primaryjoin='Customer.customer_id == CustomerLocation.customer_id')
    tbred_ship_code = Column(Unicode(6))
    tbred_bill_to = Column(Unicode(6))
    ship_method_id = Column(Integer,ForeignKey('ship_methods.ship_method_id'))
    ship_method = relation('ShipMethod',primaryjoin='ShipMethod.ship_method_id == CustomerLocation.ship_method_id')
    residential = Column(Boolean,default=False,nullable=False)
    free_shipping = Column(Boolean,default=False,nullable=False)
    collect = Column(Boolean,default=False,nullable=False)
    third_party = Column(Boolean,default=False,nullable=False)
    shipping_account = Column(Unicode(50))
    bill_to_id = Column(Integer,ForeignKey('locations.location_id'))
    bill_to = relation('CustomerLocation',remote_side=['locations.location_id'])

Answer 1

查看我的answer to a related question。通过在表中声明一个自引用外键，可以在声明中具有自引用关系，并在声明它之后对其进行“猴子修补”，或者将外部列名称指定为字符串而不是类字段。例如：

class Employee(Base):
  __tablename__ = 'employee'
  id = Column(Integer, primary_key=True)
  name = Column(String(64), nullable=False)
Employee.manager_id = Column(Integer, ForeignKey(Employee.id))
Employee.manager = relationship(Employee, backref='subordinates',
    remote_side=Employee.id)

我已成功使用此技术，之前为您提供了父子树关系的两个方向（单个父级可以有多个子记录）。如果省略backref参数，它可能适用于您，也可能不适合您。您总是可以选择在应用程序中仅使用关系的一个方向。

SQLAlchemy，同一个表上的一对一关系

1 个答案: