使用
SQLAlchemy,我与两个表 – 用户和分数有一对多的关系.我试图根据他们的总分在过去X天的天数中查询排名前10位的用户.
users: id user_name score scores: user score_amount created
我当前的查询是：
top_users = DBSession.query(User).options(eagerload('scores')).filter_by(User.scores.created > somedate).order_by(func.sum(User.scores).desc()).all()
我知道这显然不正确,这只是我最好的猜测.然而,在查看文档和谷歌搜索后,我找不到答案.
SELECT user.*,SUM(scores.amount) as score_increase FROM user LEFT JOIN scores ON scores.user_id = user.user_id WITH scores.created_at > someday ORDER BY score_increase DESC
解决方法@H_502_17@
单连接方式,在所有用户列中添加了group_by,但如果您选择：MysqL将允许您仅对“id”列进行分组：
sess.query(User,func.sum(score.amount).label('score_increase')).\
join(User.scores).\
filter(score.created_at > someday).\
group_by(User).\
order_by("score increase desc")
或者如果您只想要结果中的用户：
sess.query(User).\
join(User.scores).\
filter(score.created_at > someday).\
group_by(User).\
order_by(func.sum(score.amount))
上述两个都没有效率,因为您将所有“用户”列分组(或者您只使用MysqL的“只有几列”组,这仅是MysqL).为了最小化,子查询方法：
subq = sess.query(score.user_id,func.sum(score.amount).label('score_increase')).\
filter(score.created_at > someday).\
group_by(score.user_id).subquery()
sess.query(User).join((subq,subq.c.user_id==User.user_id)).order_by(subq.c.score_increase)
相同场景的示例在ORM教程中：http://docs.sqlalchemy.org/en/latest/orm/tutorial.html#selecting-entities-from-subqueries
sess.query(User,func.sum(score.amount).label('score_increase')).\ join(User.scores).\ filter(score.created_at > someday).\ group_by(User).\ order_by("score increase desc")
或者如果您只想要结果中的用户：
sess.query(User).\
join(User.scores).\
filter(score.created_at > someday).\
group_by(User).\
order_by(func.sum(score.amount))
上述两个都没有效率,因为您将所有“用户”列分组(或者您只使用MysqL的“只有几列”组,这仅是MysqL).为了最小化,子查询方法：
subq = sess.query(score.user_id,func.sum(score.amount).label('score_increase')).\ filter(score.created_at > someday).\ group_by(score.user_id).subquery() sess.query(User).join((subq,subq.c.user_id==User.user_id)).order_by(subq.c.score_increase)
相同场景的示例在ORM教程中：http://docs.sqlalchemy.org/en/latest/orm/tutorial.html#selecting-entities-from-subqueries