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 u.name,c.name,`date` from (select *,max(date) over (partition by user_id ) d from login) l inner join user u on u.id=l.user_id inner join client c on c.id=l.client_id where d=date order by u.name asc
虽然但是 总有分开的那一天