python - 筛选 SQLAlchemy 关系

Question

我正在使用 Pyramid + SQLAlchemy 并进行了以下设置

from sqlalchemy import orm
from __init__ import user_table  
from device import Device

class User(object):
  def __repr__(self):
    return 'User: ' + self.username

orm.mapper(User, user_table, properties={
  'devices': orm.relation(Device, backref='owner')
})

访问时如何添加过滤器user.devices？当我尝试user.devices.filter(Device.platform==1).all()时，我收到以下错误：

'InstrumentedList' object has no attribute 'filter'

Answer 1

您可以使用启用查询的属性：

class User(object):
    def __repr__(self):
        return 'User: ' + self.username

    # property
    @property
    def devices_filtered(self):
        return object_session(self).query(Device).with_parent(self).filter(Device.platform==1).all()

    # or simply a method
    def devices_filtered(self, platform):
        return object_session(self).query(Device).with_parent(self).filter(Device.platform==platform).all()

或查看动态关系加载器：

orm.mapper(User, user_table, properties={
  'devices': orm.relation(Device, backref='owner', lazy='dynamic')
})

# ...
user = session.query(User).get(_some_user_id)
devices = user.devices.filter(Device.platform==platform).all()

只需阅读参考文档以获取有关主题的完整说明。