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
为什么不创建一个table relationships?
实施后,为了获得您想要的:
其中.departments属性是关系的属性。
这也适用,但使用子选择:
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