Question

我发现如果单表继承设置中的子类中存在__table_args__，则SQLAlchemy会抛出sqlalchemy.exc.ArgumentError: Can't place __table_args__ on an inherited class with no table。同时，可以在子类上定义一个index=True的列，它以与__table_args__相同的方式更改父表。

这是我的设置：

import sqlalchemy as sa
import sqlalchemy.ext.declarative

from sqlalchemy.ext.declarative import declarative_base


Base = declarative_base()


class A(Base):
    __tablename__ = 'a'

    id = sa.Column(sa.Integer, primary_key=True)
    type = sa.Column(sa.Text, nullable=False)

    __mapper_args__ = {
        'polymorphic_on': type,
    }


class B(A):
    b = sa.Column(sa.Integer, index=True)

    __mapper_args__ = {
        'polymorphic_identity': 'b',
    }


class C(A):
    c = sa.Column(sa.Integer)

    __mapper_args__ = {
        'polymorphic_identity': 'c',
    }

    __table_args__ = (
        sa.Index('ix_test', c),
    )


engine = sa.engine.create_engine("sqlite://", echo=True)
Base.metadata.create_all(engine)
session = sa.orm.Session(engine)

session.add_all([
    B()
])

session.commit()

print(session.query(A))

它抛出：

Traceback (most recent call last):
 File "test.py", line 29, in <module>
   class C(A):
 File "/home/andrei/projects/my_project/.tox/dev/lib/python3.5/site-packages/sqlalchemy/ext/decla
rative/api.py", line 64, in __init__
   _as_declarative(cls, classname, cls.__dict__)
 File "/home/andrei/projects/my_project/.tox/dev/lib/python3.5/site-packages/sqlalchemy/ext/decla
rative/base.py", line 88, in _as_declarative
   _MapperConfig.setup_mapping(cls, classname, dict_)
 File "/home/andrei/projects/my_project/.tox/dev/lib/python3.5/site-packages/sqlalchemy/ext/decla
rative/base.py", line 103, in setup_mapping
   cfg_cls(cls_, classname, dict_)
 File "/home/andrei/projects/my_project/.tox/dev/lib/python3.5/site-packages/sqlalchemy/ext/decla
rative/base.py", line 133, in __init__
   self._setup_inheritance()
 File "/home/andrei/projects/my_project/.tox/dev/lib/python3.5/site-packages/sqlalchemy/ext/decla
rative/base.py", line 442, in _setup_inheritance
   "Can't place __table_args__ on an inherited class "
sqlalchemy.exc.ArgumentError: Can't place __table_args__ on an inherited class with no table.

有人知道任何变通方法吗？

Answer 1

指数可以放在表格定义之外。

class C(A):
    c = sa.Column(sa.Integer)

    __mapper_args__ = {
        'polymorphic_identity': 'c',
    }

Index('ix_test', C.c)

Answer 2

我在更新到SQLAlchemy 2.3后开始出现此错误。

如果您可以删除表参数，它将起作用（但可能会抱怨其他事情，特别是如果两个子类共享一个字段名称 - 这对我来说是__table_args__ = {'extend_existing': True}。

最后，解决方法是明确指定子类上的表名，例如：

class Establishment(Model):
    __tablename__ = 'establishments'
    name = Column(String(32))

class Cafe(Establishment):
    __mapper_args__ = {'polymorphic_identity': 'cafe'}
    coffee_type = Column(String(32))

class Bar(Establishment):
    __tablename__ = 'establishments'
    __table_args__ = {'extend_existing': True}
    __mapper_args__ = {'polymorphic_identity': 'bar'}
    alcohol_license_number = Column(String(40))
    bar_type = Column(Enum(BarType)) # defined elsewhere

class Restaurant(Establishment):
    __tablename__ = 'establishments'
    __table_args__ = {'extend_existing': True}
    __mapper_args__ = {'polymorphic_identity': 'bar'}
    alcohol_license_number = Column(String(40))
    cuisine_list = Column(String(255))

一个粗略的例子，但希望这个想法有意义 - 餐厅和酒吧都有一个字段来保存他们的酒类许可证信息，所以我们希望这两个人使用表中的现有字段而不是冲突 - 现在我们必须指定tablename自SQLAlchemy 2.3。

我知道这个例子与问题（关于索引）的问题不完全相同，但它会导致同样的错误，这个解决方案应该适用于这种情况。

无法在单表继承设置中的子类上定义__table_args__

2 个答案: