题解 | #牛客每个人最近的登录日期(一)#
牛客每个人最近的登录日期(一)
http://www.nowcoder.com/practice/ca274ebe6eac40ab9c33ced3f2223bb2
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;
