使用SQLAlchemy v1.4向继承表添加行时遇非空列NULL错误

为演示需求，涉及四张表：ModelType、ModelTypeA、ModelTypeB、Model，需通过SQLAlchemy v1.4实现指定的表继承关联关系。已通过以下代码定义实体类并成功创建表结构：

Base = declarative_base()

class ModelType(Base):
    __tablename__ = "modeltype"

    id = Column(Integer, primary_key=True)
    algorithm = Column(String)

    models = relationship("Model", back_populates="modeltype")

    __mapper_args__ = {
        "polymorphic_identity": "modeltype",
        "polymorphic_on": algorithm
    }

    def __repr__(self):
        return f"{self.__class__.__name__}({self.algorithm!r})"

class ModelTypeA(ModelType):
    __tablename__ = "modeltypea"

    id = Column(Integer, ForeignKey("modeltype.id"), primary_key=True)
    parameter_a = Column(Integer)

    __mapper_args__ = {
        "polymorphic_identity": "Model Type A"
    }
    
class ModelTypeB(ModelType):
    __tablename__ = "modeltypeb"

    id = Column(Integer, ForeignKey("modeltype.id"), primary_key=True)
    parameter_a = Column(Integer)

    __mapper_args__ = {
        "polymorphic_identity": "Model Type B"
    }

class Model(Base):
    __tablename__ = "model"

    id = Column(Integer, primary_key=True)
    trainingtime = Column(Integer)
    modelversionid = Column(Integer, ForeignKey("modeltype.id"))
    
    modeltype = relationship("ModelType", back_populates="models")

    def __repr__(self) -> str:
        return f"Model(id={self.id!r}, trainingtime={self.trainingtime!r})"

创建表的代码及生成的SQL语句均正常，但执行以下添加入行操作时：

model_type_a = ModelTypeA(parameter_a=3)

model = Model(trainingtime=10, modeltype=model_type_a)

session.add(model)
session.commit()

出现SAWarning警告：列'modeltypea.id'作为主键无默认值且未传值，随后触发sqlalchemy.exc.IntegrityError错误：非空列出现NULL结果。推测创建ModelTypeA实例时未在ModelType表生成对应行，导致无可用id关联，请问该继承表场景下应如何正确添加入行？

这个问题出在SQLAlchemyjoined-table继承的配置上，子类表的主键需要和父类主键自动同步值，只需修改子类的id字段配置即可解决：

• 给子类的id字段添加autoincrement=False参数，明确该字段的值完全依赖父类主键生成，不需要自身自增逻辑

修正后的子类代码如下：

class ModelTypeA(ModelType):
    __tablename__ = "modeltypea"

    # 添加autoincrement=False，指定字段值由父类主键同步
    id = Column(Integer, ForeignKey("modeltype.id"), primary_key=True, autoincrement=False)
    parameter_a = Column(Integer)

    __mapper_args__ = {
        "polymorphic_identity": "Model Type A"
    }
    
class ModelTypeB(ModelType):
    __tablename__ = "modeltypeb"

    id = Column(Integer, ForeignKey("modeltype.id"), primary_key=True, autoincrement=False)
    parameter_a = Column(Integer)

    __mapper_args__ = {
        "polymorphic_identity": "Model Type B"
    }

修改后原插入代码无需调整，执行session.commit()时，SQLAlchemy会自动完成以下流程：

1. 先向ModelType表插入数据，生成主键id
2. 将该主键值同步到ModelTypeA表的id字段，完成子类数据插入
3. 最后插入Model表数据并关联对应的外键值

内容的提问来源于stack exchange，提问作者Luca Guarro