题解 | 牛客每个人最近的登录日期(二)
牛客每个人最近的登录日期(二)
https://www.nowcoder.com/practice/7cc3c814329546e89e71bb45c805c9ad
select user_name as u_n, client_name as c_n, date from( select t1.*, t2.name as user_name, t3.name as client_name , rank() over(partition by t1.user_id order by t1.date desc) as ranking from login as t1 left join user as t2 on t1.user_id = t2.id left join client as t3 on t1.client_id = t3.id) as t4 where ranking = 1 order by u_n;
