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)
http://www.nowcoder.com/practice/e524dc7450234395aa21c75303a42b0a
解题思路:
将每个用户的首次登录时间生成一张临时表,然后与现在这张表右连接
select date, count(if(s1.first_day = l.date,true,null))
from
(
select user_id, min(date) as first_day
from login
group by user_id
)s1
right join login l
on s1.user_id = l.user_id
group by l.date
;