题解 | #牛客每个人最近的登录日期(二)#

写的时候几个子查询嵌套给我整昏了，但是逻辑还是通顺的，就是括号太多了...

select temp3.name, client.name, date from 

(select user.name, client_id, date from 
    
        (select user_id, client_id, date from 
            (
                select *, row_number() over (partition by user_id 
                        order by date  DESC) as rn 
                            from login
            ) temp1
                where rn = 1
        ) temp2 
            join user on user_id = user.id
    ) temp3 
        join client on client_id = client.id 
order by temp3.name