题解 | #牛客每个人最近的登录日期(二)#

        ,c_n
        ,date
from
(
select   tb_2.name as u_n
        ,tb_3.name as c_n
        ,tb_1.date
        ,row_number() over (partition by tb_2.name order by date desc) as max_date
from
(
(
    select   id
            ,user_id
            ,client_id
            ,date
    from login
) tb_1
left join
(
    select   id
            ,name
    from user
) tb_2
on tb_1.user_id = tb_2.id

left join
(
    select   id
            ,name
    from client
) tb_3
on tb_1.client_id = tb_3.id
)
) t
where max_date = 1