我希望以下列方式获得属于父类的对象列表：

class A(object):
    __tablename__ = 'a'
    id = Column(Integer, primary_key=True)
    collection = relationship(.....) # contains an ordered list of [B, C, B, B, C, C, C, B, C, C, ...];

class B(object):
    __tablename__ = 'b'
    id = Column(Integer, primary_key=True)

class C(object):
    __tablename__ = 'c'
    id = Column(Integer, primary_key=True)

SQLAlchemy examples文件夹有一个简单的多对一,在我的例子中,类B和C是A的“父”类(而不是相反),但我不能为我的生活做出努力如何将其转换为一对多,然后添加双向关系,使其成为多对多关系.

任何人都可以帮我解决这个问题吗？

解决方法:

它有点痛苦,而AbstractConcreteBase显然需要更多的润色,但它是这样的：

from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base, AbstractConcreteBase

Base = declarative_base()

class A(Base):
    __tablename__ = 'a'
    id = Column(Integer, primary_key=True)

class BC(AbstractConcreteBase, Base):
    pass

class B(BC):
    __tablename__ = 'b'
    id = Column(Integer, primary_key=True)

    a_id = Column(Integer, ForeignKey('a.id'))
    __mapper_args__ = {
        "polymorphic_identity": "b",
        "concrete": True
    }

class C(BC):
    __tablename__ = 'c'
    id = Column(Integer, primary_key=True)
    a_id = Column(Integer, ForeignKey('a.id'))
    __mapper_args__ = {
        "polymorphic_identity": "c",
        "concrete": True
    }

configure_mappers()
A.collection = relationship(BC, primaryjoin=BC.a_id == A.id)

engine = create_engine("sqlite://", echo=True)

Base.metadata.create_all(engine)

sess = Session(engine)

sess.add_all([
    A(collection=[
        B(),
        C(),
        C()
    ]),
    A(collection=[
        B(),
        B()
    ])
])

sess.commit()

for a in sess.query(A):
    for bc in a.collection:
        print a, bc