Question

我在SQLAlchemy的Flask应用中使用关联对象设置了多对多的关系。然后，我在类之间设置了关联代理，以提供更直接的访问，而不是通过关联对象。

以下是设置的缩写示例：

class Person(Model):
    __tablename__ = 'persons'
    id = Column(Integer, primary_key=True)
    last_name = Column(Text, nullable=False)
    groups = association_proxy('group_memberships', 'group')
    # Other stuff

class Group(Model):
    __tablename__ = 'groups'
    id = Column(Integer, primary_key=True)
    name = Column(Text, nullable=False)
    members = association_proxy('group_memberships', 'person')
    # Other stuff

class GroupMembership(Model):
    __tablename__ = 'group_memberships'
    id = Column(Integer, primary_key=True)
    person_id = Column(Integer, ForeignKey('persons.id'), nullable=False)
    group_id = Column(Integer, ForeignKey('groups.id'), nullable=False)
    person = relationship('Person', uselist=False, backref=backref('group_memberships', cascade='all, delete-orphan'))
    group = relationship('Group', uselist=False, backref=backref('group_memberships', cascade='all, delete-orphan'))    
    # Other stuff

我不能为我的生活弄清楚如何让group.members返回的成员按其last_name排序？

Answer 1

为了对group.members进行排序，您必须在加载GroupMembership关联对象时让人员可以进行排序。这可以通过连接来实现。

在当前配置中，访问group.members首先加载GroupMembership对象，填充group.group_memberships关系，然后在关联代理访问GroupMembership.person关系属性时为每个Person触发SELECT。 / p>

相反，您希望在同一查询中加载GroupMemberships和Persons，按Person.last_name排序：

class GroupMembership(Model):
    __tablename__ = 'group_memberships'
    id = Column(Integer, primary_key=True)
    person_id = Column(Integer, ForeignKey('persons.id'), nullable=False)
    group_id = Column(Integer, ForeignKey('groups.id'), nullable=False)
    person = relationship('Person',
                          backref=backref('group_memberships',
                                          cascade='all, delete-orphan'),
                          lazy='joined', innerjoin=True,
                          order_by='Person.last_name')
    group = relationship('Group', backref=backref('group_memberships',
                                                  cascade='all, delete-orphan'))
    # Other stuff

您需要在标量关系属性order_by='Person.last_name'上定义GroupMembership.person，而不是backref Group.group_memberships，这看似合乎逻辑。另一方面，order_by“表示加载这些项时应该应用的顺序”，因此在使用joined loading时有意义。由于您将加入多对一引用并且外键不可为空，因此您可以使用内部联接。

根据给定的定义：

In [5]: g = Group(name='The Group')

In [6]: session.add_all([GroupMembership(person=Person(last_name=str(i)), group=g)
   ...:                  for i in range(30, 20, -1)])

In [7]: session.commit()

In [8]: g.members
2017-06-29 09:17:37,652 INFO sqlalchemy.engine.base.Engine BEGIN (implicit)
2017-06-29 09:17:37,653 INFO sqlalchemy.engine.base.Engine SELECT groups.id AS groups_id, groups.name AS groups_name 
FROM groups 
WHERE groups.id = ?
2017-06-29 09:17:37,653 INFO sqlalchemy.engine.base.Engine (1,)
2017-06-29 09:17:37,655 INFO sqlalchemy.engine.base.Engine SELECT group_memberships.id AS group_memberships_id, group_memberships.person_id AS group_memberships_person_id, group_memberships.group_id AS group_memberships_group_id, persons_1.id AS persons_1_id, persons_1.last_name AS persons_1_last_name 
FROM group_memberships JOIN persons AS persons_1 ON persons_1.id = group_memberships.person_id 
WHERE ? = group_memberships.group_id ORDER BY persons_1.last_name
2017-06-29 09:17:37,655 INFO sqlalchemy.engine.base.Engine (1,)
Out[8]: [<__main__.Person object at 0x7f8f014bdac8>, <__main__.Person object at 0x7f8f014bdba8>, <__main__.Person object at 0x7f8f014bdc88>, <__main__.Person object at 0x7f8f01ddc390>, <__main__.Person object at 0x7f8f01ddc048>, <__main__.Person object at 0x7f8f014bdd30>, <__main__.Person object at 0x7f8f014bde10>, <__main__.Person object at 0x7f8f014bdef0>, <__main__.Person object at 0x7f8f014bdfd0>, <__main__.Person object at 0x7f8f0143b0f0>]

In [9]: [p.last_name for p in _]
Out[9]: ['21', '22', '23', '24', '25', '26', '27', '28', '29', '30']

此解决方案的缺点是person关系总是急切地加载，并且在查询GroupMemberships时应用了ORDER BY：

In [11]: session.query(GroupMembership).all()
2017-06-29 12:33:28,578 INFO sqlalchemy.engine.base.Engine BEGIN (implicit)
2017-06-29 12:33:28,578 INFO sqlalchemy.engine.base.Engine SELECT group_memberships.id AS group_memberships_id, group_memberships.person_id AS group_memberships_person_id, group_memberships.group_id AS group_memberships_group_id, persons_1.id AS persons_1_id, persons_1.last_name AS persons_1_last_name 
FROM group_memberships JOIN persons AS persons_1 ON persons_1.id = group_memberships.person_id ORDER BY persons_1.last_name
2017-06-29 12:33:28,578 INFO sqlalchemy.engine.base.Engine ()
Out[11]: 
    ...

...除非明确使用其他加载策略：

In [16]: session.query(GroupMembership).options(lazyload('person')).all()
2018-04-05 21:10:52,404 INFO sqlalchemy.engine.base.Engine BEGIN (implicit)
2018-04-05 21:10:52,405 INFO sqlalchemy.engine.base.Engine SELECT group_memberships.id AS group_memberships_id, group_memberships.person_id AS group_memberships_person_id, group_memberships.group_id AS group_memberships_group_id 
FROM group_memberships
2018-04-05 21:10:52,405 INFO sqlalchemy.engine.base.Engine ()

如果您不时需要备用订单，则必须恢复发出使用explicit eager loading的完整查询并按以下顺序排序：

In [42]: g = session.query(Group).\
    ...:     filter_by(id=1).\
    ...:     join(GroupMembership).\
    ...:     join(Person).\
    ...:     options(contains_eager('group_memberships')
    ...:             .contains_eager('person')).\
    ...:     order_by(Person.last_name.desc()).\
    ...:     one()
    ...:             

In [43]: [m.last_name for m in g.members]
Out[43]: ['30', '29', '28', '27', '26', '25', '24', '23', '22', '21']

SQLAlchemy order_by通过关联代理的多对多关系

1 个答案: