题解 | #牛客的课程订单分析(五)#

select *
from(
select 
distinct
user_id,
min(`date`)over(partition by user_id) as first_buy_date,
if(dense_rank()over(partition by user_id order by date asc) = 2,`date`,null) as second_buy_date,
count(*)over(partition by user_id) as cnt
from order_info
where `date` > '2025-10-15'
and product_name in ('C++','Java','Python')
and status = 'completed'
) as t1
where t1.second_buy_date is not null;