题解 | #牛客每个人最近的登录日期(二)#

select 
    t2.name u_n,
    t3.name  c_n,
    t1.date
from 
    login t1
    join user t2 on t1.user_id = t2.id
    join client t3 on t1.client_id = t3.id
    join (select id, max(date) over (partition by user_id ) max_date from login)  t4
    on t1.id = t4.id
where 
    t1.date = t4.max_date
order by
    t2.name