题解 | #牛客每个人最近的登录日期(六)#
牛客每个人最近的登录日期(六)
http://www.nowcoder.com/practice/572a027e52804c058e1f8b0c5e8a65b4
问题描述:请你写出一个sql语句查询刷题信息,包括: 用户的名字,以及截止到某天,累计总共通过了多少题,并且查询结果先按照日期升序排序,再按照姓名升序排序,有登录却没有刷题的哪一天的数据不需要输出。
方案1:联结passing_number,user表+窗口函数
SELECT u.name as u_n,date,sum(p.number) over (partition by p.user_id order by p.date ASC) AS ps_num FROM passing_number p INNER JOIN user u ON p.user_id = u.id GROUP BY p.date,p.user_id ORDER BY p.date,u.name
方案2:passing_number笛卡尔自连接+case 联结user表
select u.name as u_n,p1.date,sum(case when p1.date>=p2.date then p2.number else 0 end) as ps_num FROM passing_number p1 cross join passing_number p2 ON p1.user_id=p2.user_id inner join user u on p1.user_id = u.id group by p1.user_id,p1.date order by p1.date,u_n;
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