sql面试题：求连续最大登录天数

求用户最大连续登陆天数mysql实现

with t_1 as( -- 使用timestampdiff(unit, begin, end) <= 30进行where的条件筛选，并使用distinct函数进行清洗去重得到新表t_1
select DISTINCT  dimMemberID as user_id, date(dimDateID) as login_date
from fct_sales
where timestampdiff(day, date(dimDateID),(select max(dimDateID) from fct_sales)) <=  30
      and  dimMemberID <> 0
order by login_date
),t_2 as( -- 得到一个按user_id分组并有排序编号的新表t_2
select  user_id
       ,login_date
       ,row_number()over(partition by user_id order by login_date asc ) as ranks
from t_1
),t_3 as ( -- 使用date_sub函数获得新的列gap和新表t_3
select   user_id 
		,login_date
		,ranks
		,date_sub(login_date, interval ranks day) as gap
from t_2
),  t_4 as ( -- 对t_3表按user_id、gap进行分组，并对组内的login_date进行count计数，得到表t_4
select  user_id
       ,gap
       ,count(login_date) as num
from t_3
group by  user_id
         ,gap
) ,t_5 as ( -- 按照user_id进行分组，使用max取出最大的num
select  user_id
       ,max(num) as series_login -- 按照user_id分组，求得最大连续登陆天数
from t_4 
group by user_id
)
select * ,count(1)over() as cnt  from t_5;

• with t_1 as
• timestampdiff

参考