使用SQLAlchemy进行多对多表的多次连接

2024-09-28 15:32:09 发布

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SQLalchemy的新特性,这里是我的问题:

我的模型是:

user_group_association_table = Table('user_group_association', Base.metadata,
    Column('user_id', Integer, ForeignKey('user.id')), 
    Column('group_id', Integer, ForeignKey('group.id'))    
)

department_group_association_table = Table('department_group_association', Base.metadata,
    Column('department', Integer, ForeignKey('department.id')), 
    Column('group_id', Integer, ForeignKey('group.id'))
)

class Department(Base):
    __tablename__ = 'department'
    id = Column(Integer, primary_key=True)
    name = Column(String(50))


class Group(Base):
    __tablename__ = 'group'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    users = relationship("User", secondary=user_group_association_table, backref="groups")
    departments = relationship("Department", secondary=department_group_association_table, backref="groups")

class User(Base):

    __tablename__ = 'user'
    id = Column(Integer, primary_key=True)
    firstname = Column(String(50))
    surname = Column(String(50))

因此,此代码反映了以下关系:

   --------             ---------             --------------
   | User | --- N:M --- | Group | --- N:M --- | Department |
   --------             ---------             --------------

我尝试使用joins,但仍未能成功执行以下操作:

一个sqlalchemy请求获取所有用户实例,同时知道部门名称(例如“R&D”)

首先应该:

session.query(User).join(...
or
session.query(User).options(joinedLoad(...

有人能帮忙吗?

谢谢你的时间

皮埃尔


Tags: idbasestringtablegroupcolumnintegerclass
2条回答

为什么不创建一个table relationships

实施后,为了获得您想要的:

list_of_rnd_users = [u for u in Users if 'R&D' in u.departments]

其中.departments属性是关系的属性。

session.query(User).join((Group, User.groups)) \
    .join((Department, Group.departments)).filter(Department.name == 'R&D')

这也适用,但使用子选择:

session.query(User).join((Group, User.groups)) \
    .filter(Group.departments.any(Department.name == 'R&D'))

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