Python SqlAlchemy链接表

所以我想用SQLAlchemy创建3个表

Obj
obj_id | obj_type

User
user_id | obj_id | name | mail 

Society
society_id | obj_id | name 

让我们假设我推3个元素，我想要这样的东西：

Obj
obj_id | obj_type
1      | society
2      | user
3      | society

User
user_id | obj_id | name | mail 
1       | 2      | John | john.smith@700.com

Society
society_id | obj_id | name
1          | 1      | Google
2          | 3      | Facebook

如何构造代码并添加/提交以便链接表

class Obj(Base):
    __tablename__ = 'objects'
    obj_id = Column(Integer, primary_key=True)
    obj_type = Column(String(30))

class User(Base):
    __tablename__ = 'users'
    user_id = Column(Integer, primary_key=True)
    obj_id = Column(Integer, ForeignKey('objects.obj_id'))
    name = Column(String(100), default=None)
    mail = Column(String(200), default=None)
    objects = relationship("Obj", back_populates="users")

class Society(Base):
    __tablename__ = 'society'
    society_id = Column(Integer, primary_key=True)
    obj_id = Column(Integer, ForeignKey("objects.obj_id"))
    name = Column(String(100), default=None)
    objects = relationship("Obj", back_populates="society")
    people = relationship("People", back_populates="society")

---

您的表结构类似于，其中

objects.obj_type

为：

请注意，您不需要子类中的对象关系来将它们链接到相关的Obj。您可以像继承时一样访问父对象的属性

添加3个元素：

In [6]: session.add(User(name='John', mail='john.smith@700.com'))

In [7]: session.add_all([Society(name='Google'), Society(name='Facebook')])

In [8]: session.commit()

并检查是否可以查询所有OBJ：

In [9]: session.query(Obj).all()
Out[9]: 
[<__main__.User at 0x7f37b5aaa780>,
 <__main__.Society at 0x7f37b5aaab00>,
 <__main__.Society at 0x7f37b5aaacc0>]

In [9]: session.query(Obj).all()
Out[9]: 
[<__main__.User at 0x7f37b5aaa780>,
 <__main__.Society at 0x7f37b5aaab00>,
 <__main__.Society at 0x7f37b5aaacc0>]

In [10]: session.query(Obj.__table__).all()
Out[10]: [(1, 'user'), (2, 'society'), (3, 'society')]

In [11]: session.query(User.__table__).all()
Out[11]: [(1, 1, 'John', 'john.smith@700.com')]

In [12]: session.query(Society.__table__).all()
Out[12]: [(1, 2, 'Google'), (2, 3, 'Facebook')]