select distinct user_id,count(1) as days_cnt
from(
select *,row_number() over (partition by user_id order by sales_date) as rk from sales_tb) as t1
group by user_id,date_sub(sales_date,interval rk day)
having count(1) >= 2

按照user_id分组使用窗口函数，如果天数连续，则sales_date - rk应该是相同的，再按照user_id和sales_date - rk分组，拿到最终的不同用户连续登录天数的信息