经典的求连续天数问题

统一的思路都是先按天数进行排序，再用天数减去排名进行count得到连续登录的天数

with t as (select 
    distinct user_id,
    sales_date,
    dense_rank() over(partition by user_id order by sales_date) rk
from sales_tb)

select 
    user_id,
    count(sales_date - rk)
from t
group by user_id
having count(sales_date - rk) >= 2
order by user_id