sqlalchemy支持单表继承。
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,我会发现错误:
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'为表属性时:
__table_args__ = {'extend_existing': True}
然后,当我尝试使用任何模型的任何内容时,我会得到下一个错误:
sqlalchemy.exc.NoForeignKeysError: Can't find any foreign key relationships between 'tbl_user' and 'tbl_user'.
这应该是直接的,应该可以通过单个属性来解决,尤其是与Sqlalchemy效果很好。有什么想法解决问题?
我终于找到了一种提及应该如何工作的方式。我需要从初始mapper args中删除 with_polymorfic
选项,并为儿童课程添加 __tablename__ = None
,因此准确的代码看起来像:
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'
}
...