题解 | #牛客每个人最近的登录日期(四)#
牛客每个人最近的登录日期(四)
https://www.nowcoder.com/practice/e524dc7450234395aa21c75303a42b0a
比较简洁的写法 select date,sum(ne) new from (select *,if(first_value(date)over(partition by user_id order by date)=date,1,0) ne from login) l1 group by date;