sql获取连续登录不少于7天的用户

测试数据:

 解题思路：

1.用窗口函数根据userId分组，将每个分组内的数据根据排序顺序编号

select
    userid,
    loginDate ,
    ROW_NUMBER() OVER (PARTITION BY userId ORDER BY loginDate) AS seq
	from test;

2. 用每条记录的loginDate 减去seq，得到一个时间 ，如果两条记录得到的这个时间相等表示是连续登录

select
    userid,
    loginDate ,
    ROW_NUMBER() OVER (PARTITION BY userId ORDER BY loginDate) AS seq,
    date_sub(logindate,interval row_number() over (partition by userid order by logindate) day) as day
	from test;

3.根据题目要求统计连续登录大于等于7天的用户，就是统计同一个userid是否至少存在7条记录的day是相等的

with gr as 
(select
	userid,
	date_sub(logindate,interval row_number() over (partition by userid order by logindate) day) as day
	from test)
select
    userid,
    day,
    count(1)
from gr group by userId ,day HAVING (count(1)>=7);

 上面的写法等同于

select
    userid,
    day,
    count(1)
from 
(select
	userid,
	date_sub(logindate,interval row_number() over (partition by userid order by logindate) day) as day
	from test) t
group by userId ,day HAVING (count(1)>=7);