题解 | #牛客每个人最近的登录日期(五)#
牛客每个人最近的登录日期(五)
http://www.nowcoder.com/practice/ea0c56cd700344b590182aad03cc61b8
基于牛客每个人最近的登录日期(三)的题解
select distinct login.date,ifnull(b.p,0) from login left join ( select a.first_date,round(count(l.date)/count(*),3) as p from (select user_id, min(date) as first_date from login GROUP by user_id) as a left join login l on a.user_id=l.user_id and l.date=DATE_ADD(a.first_date,INTERVAL 1 DAY) group by a.first_date )b # 有新用户登录的日期，及其次日留存率 on login.date=b.first_date order by login.date
