python - 从父类(super class)实例化子类

Question

我有以下 python 代码，它使用 SQLAlchemy 在 MySQL 数据库中创建多态结构。

class Animal(Base):
    key = Column(Integer(), Sequence("My Counter" ,1 ,1), primary_key = True)
    name = Column(String())
    discriminator = Column('type',String())
    __mapper_args__ = {'polymorphic_on':discriminator}

    def __init__(self,key,name):
        self.key = key
        self.name = name

class Cat(Animal):
    __tablename__ = 'cat'
    __mapper_args__ = {'polymorphic_identity':'cat'}
    def walk():
        pass

class Dog(Animal):
    __tablename__ = 'dog'
    __mapper_args__ = {'polymorphic_identity':'dog'}

    def walk():
         pass

我想知道加载它的最佳方法是什么:

a = Animal(key=1)
c = a.create()

c 现在将是 Cat 或 Dog 对象，具体取决于它的类型。 Animal 表包含此信息。

谢谢

Answer 1

这不是使用 sqlalchemy 查询数据库的方式。为了获得特定 key 的动物，您需要做的就是执行:

my_key = 1
a = session.query(Animal).get(my_key)
# sqlalchemy will figure out the type automatically and will return object of proper class 
assert type(a) in (Cat, Dog,)

请注意，您提供的模型不完整。下面的模型应该是完整的工作模型:

class Animal(Base):
    __tablename__ = 'animal'
    key = Column(Integer(), Sequence("My Counter" ,1 ,1), primary_key = True)
    name = Column(String())
    discriminator = Column('type',String())
    __mapper_args__ = { 'polymorphic_on':discriminator, }

    def __init__(self,key,name):
        self.key = key
        self.name = name

class Cat(Animal):
    __tablename__ = 'cat'
    __mapper_args__ = {'polymorphic_identity':'cat'}
    key = Column(Integer(), ForeignKey('animal.key'), primary_key = True)

    def walk(self):
        print "cat walking"

class Dog(Animal):
    __tablename__ = 'dog'
    __mapper_args__ = {'polymorphic_identity':'dog'}
    key = Column(Integer(), ForeignKey('animal.key'), primary_key = True)

    def walk(self):
        print "dog walking"