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 b.name as u_n, c.name as c_n, a.date
from (
select user_id, client_id, date, row_number() over(partition by user_id order by date desc) as ranks from login
) a
join user b on a.user_id = b.id
join client c on c.id = a.client_id
where a.ranks = 1
order by u_n asc
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