题解 | #牛客每个人最近的登录日期(二)#

select u_n,
       c_n,
       date
from (
select b.name u_n,
       c.name c_n, 
       date, RANK() OVER (PARTITION BY b.name ORDER BY date DESC) AS user_order
from login a
left join user b on a.user_id = b.id
left join client c on a.client_id = c.id
order by u_n
) k
where user_order =1