题解 | #牛客的课程订单分析(五)#

自己感觉最简单的一版了
select user_id,min(fir),min(sec),count(user_id) cnt
from (
    select user_id,date,fir,if(r1=2,date,null) sec
    from (
        select user_id,date,min(date)over(partition by user_id) fir,row_number()over(partition by user_id order by date) r1
        from (
            select user_id,date,if(product_name in('C++','Java','Python') and status='completed',1,0) r2
            from order_info) t3
        where r2=1 and date>'2025-10-15') t1
    ) t2
group by user_id
having count(user_id)>=2
order by user_id;