题解 | #牛客的课程订单分析(七)#

我发现模块化以后写的很快
with o as(select id,user_id,client_id,is_group_buy
from (
    select id,user_id,client_id,is_group_buy,count(id)over(partition by user_id) ding
    from order_info
    where product_name in('C++','Java','Python') and status = 'completed' and date>'2025-10-15') t1
where ding>=2
) -------这部分前面的题通用
select source,count(source) cnt
from (select if(is_group_buy='Yes','GroupBuy',name) source
from o left join client c on o.client_id=c.id
) sou
group by source
order by source;