select user_id,count(user_id) days_count
from 
(select *,sales_date-row_number() over(partition by user_id order by sales_date) sss /*连续登录天数通用,相减相等个数即连续登录天数*/
from 
(select distinct user_id,sales_date
from sales_tb) cb) cb2
group by user_id,sss
having days_count >= 2
order by user_id