1.解法一,对user_id 分组,去日期最大值 max()
select user_id,max(date) as d from login group by user_id order by user_id;2.解法二,对user_id 分组,用窗口函数对日期进行倒序排序后,取排在第一位的值
select user_id,d from (select user_id,date as d, rank() over(partition by user_id order by date desc) as rnk from login) as a where rnk=1;