题解 | #牛客每个人最近的登录日期(一)#

方案1：MAX函数+group by

SELECT user_id,MAX(date) AS d
FROM login
GROUP BY user_id
ORDER BY user_id ASC;

方案2：dense_rank窗口函数

SELECT t.user_id,t.date
FROM (SELECT *,DENSE_RANK() OVER(PARTITION BY user_id ORDER BY date DESC) AS rank_date
    FROM login) t
WHERE t.rank_date = 1
ORDER BY t.user_id;