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
with temp_table as (
select u.name as u_n, c.name as c_n, max(l.date) as latest_date
from login l
join user u on l.user_id = u.id
join client c on l.client_id = c.id
group by u.name, c.name
)
select a.u_n, a.c_n, a.latest_date from(
select *,
dense_rank() OVER (PARTITION BY u_n order by latest_date desc) as desnse_rank
from temp_table
) a where a.desnse_rank = 1