题解 | #牛客的课程订单分析(五)#可读性好就是麻烦

with first as  (
  select user_id,min(date) first_date,count(*) cnt
  from order_info
  where date>='2025-10-15'
  and status='completed'
  and product_name in ('C++','Java','Python')
  group by user_id
  having count(*)>=2
),
second as (
SELECT user_id,date second_date
from (
  select *
  ,row_number()over(partition by user_id order by date) posn
  from order_info
  where date>='2025-10-15'
  and status='completed'
  and product_name in ('C++','Java','Python')
) as t
where posn=2
)
SELECT  first.user_id,first.first_date,second.second_date,first.cnt
from first 
inner join second
on first.user_id=second.user_id
order by first.user_id