题解 | #牛客每个人最近的登录日期(五)#

select 
	date,
	round(ifnull(count(if(datediff(A.date_1,A.date) = 1,1,null))/count(if(A.num = 1,1,null)),0),3)  
from(
    select 
        user_id,
        date,
        lead(date,1) over(partition by user_id order by date asc) as date_1,
        rank() over(partition by user_id order by date asc) as num
    from login
) as A group by date

理论上来说是最简单的一种方法了 不需要链接，mysql链接也麻烦