python - Flask-SQLAlchemy 单表继承

Question

SQLAlchemy 支持 single table inheritance .

我的结构如下:

class User(db.Model):
    __tablename__ = 'tbl_user'
    type = db.Column(db.String(32)) 
    ...
    __mapper_args__ = {
        'polymorphic_identity': 'user',
        'polymorphic_on': type,
        'with_polymorphic': '*'
    }

class Tourist(User):
    __mapper_args__ = {
        'polymorphic_identity': 'tourist'
    }
    ...

class Guide(User):
    __mapper_args__ = {
        'polymorphic_identity': 'guide'
    }
    ...

当我尝试运行代码时，出现如下错误:

sqlalchemy.exc.InvalidRequestError: Table 'tbl_user' is already defined for this MetaData instance.  Specify 'extend_existing=True' to redefine options and columns on an existing Table object.

当我添加“extend_existing”作为表属性时:

__table_args__ = {'extend_existing': True}

然后当我尝试使用提到的模型时出现下一个错误:

sqlalchemy.exc.NoForeignKeysError: Can't find any foreign key relationships between 'tbl_user' and 'tbl_user'.

这应该是直截了当的事情，并且应该可以通过单个属性解决，尤其是它可以与 SQLAlchemy 一起正常工作。有解决问题的想法吗？

Answer 1

我终于找到了一种应该如何工作的方法。我需要从初始映射器参数中删除 with_polymorfic 选项，并为子类添加 __tablename__ = None，因此确切的代码如下所示:

class User(db.Model):
    __tablename__ = 'tbl_user'
    type = db.Column(db.String(32)) 
    ...
    __mapper_args__ = {
        'polymorphic_identity': 'user',
        'polymorphic_on': type,
    }  # remove with_polymorphic

class Tourist(User):
    __tablename__ = None  # Add table name to be None
    __mapper_args__ = {
        'polymorphic_identity': 'tourist'
    }
    ...

class Guide(User):
    __tablename__ = None  # Add table name to be None
    __mapper_args__ = {
        'polymorphic_identity': 'guide'
    }
    ...