Question

我有一个模型，它是抽象类的多态性身份

class AbstractModel(Base):
    type = Column(String())
    __mapper_args__ = {"polymorphic_on": type}

class ModelA(AbstractModel):
    __mapper_args__ = {"polymorphic_identity": "model_a"}

class FlaskModel(ModelA):
    __mapper_args__ = {"polymorphic_identity": "model_a"}

我需要FlaskModel与ModelA具有相同的多态关系，因为FlaskModel具有在ModelA中不存在的特定于约束的约束（请求上下文，用户权限等）

但是，在创建第二个类时，SQLAlchemy会抛出有关重复的警告，这是有充分理由的，因为任何查询始终指向FlaskModel，即使它们是从ModelA查询的。

有关完成此操作的任何建议吗？将代码拆分为包然后导入不是一个选项。

Answer 1

I ended up using this route, which worked well and allowed me to create flask specific functions.

class AbstractModel(Base):
    type = Column(String())
    __mapper_args__ = {"polymorphic_on": type}

class ModelA(AbstractModel):
    __mapper_args__ = {"polymorphic_identity": "model_a"}

class FlaskModel(ModelA):
    pass

##Event listeners are requireed for the inherited polymorphic relationships
##If additional work needs to be done with different aspect, including after_delete and after_update
##They can als be added here
@event.listens_for(FlaskModel, 'init')
def receive_label_init(target, args, kwargs):
    kwargs["type"] = "model_a"


@event.listens_for(FlaskModel, 'mapper_configured')
def receive_label_mapper_configured(mapper, class_):
    mapper.polymorphic_identity = "model_a"

不同模型的相同多态身份

1 个答案: