题解 | #牛客每个人最近的登录日期(二)#

select
    u_n,
    c_n,
    `date`
from
    (
        select
            t3.name as u_n,
            t1.name as c_n,
            t2.date as `date`,
            rank() over (
                partition by
                    t2.user_id
                order by
                    t2.date desc
            ) as ranking
        from
            client as t1
            inner join login as t2 on t1.id = t2.client_id
            inner join user as t3 on t2.user_id = t3.id
        group by
            t3.id,
            u_n,
            c_n,
            `date`
    ) as tp
where ranking = 1
order by u_n asc;