从'middle'开始的 SQLA lchemy 查询联接路径

Question

我有以下型号（用于预订服务）

from sqlalchemy_utils import ArrowType
class User:
    id = db.Column(db.Integer, primary_key=True)
    name = Column(db.String(80))
class Service:
    desc = Column(db.String(100))
    provider_id = Column(db.Integer, db.ForeignKey('user.id')
    provider = relationship('User', backref='services')
class TimeSlot:
    start = Column(ArrowType)
    owner_id = Column(db.Integer, db.ForeignKey('user.id')
    owner = relationship('User', backref ='timeslots')

我正在尝试执行加入，从用户开始，

users = (
    db.session.query(User)
        .join(Service)
        .join(TimeSlot)
        .options(contains_eager(User.services, User.timeslots)
                 )
        .filter(Service.desc == 'Fun')
        .filter(TimeSlot.start == arrow.get('2015-06-15T19:00:00+00:00'))
        .all()
    )
# would like to list the results in a 'nested' way.
for u in users:
    print(u)
    for ts in u.timeslots:
       print(' ', ts)
    for svc in u.services:
       print(' ', svc)

然而，我得到了以下错误，

属性"User.timeslots"未从元素链接'映射器|服务|服务'

我也理解这是因为.join（）和thaat Service和TimeSlot的顺序没有连接。

我应该如何获得用户及其时隙+服务的列表？

感谢

Answer 1

users = (
db.session.query(User)
    .join(Service)
    .join(TimeSlot)
    .options(contains_eager(User.services), contains_eager(User.timeslots))
    .filter(Service.desc == 'Fun')
    .filter(TimeSlot.start == arrow.get('2015-06-15T19:00:00+00:00'))
    .all()
)

Answer 2

tryfrom_joinpoint:

q = session.query(Node).
    join("children", aliased=True).
    filter(Node.name='child 1').
    join("children", aliased=True, from_joinpoint=True).
    filter(Node.name == 'grandchild 1')

或reset_joinpoint:

http://docs.sqlalchemy.org/en/latest/orm/query.html#sqlalchemy.orm.query.Query.reset_joinpoint