题解 | #牛客的课程订单分析(七)#

select source,sum(cnt) cnt
from (
select user_id,source,cnt,sum(cnt)over(partition by user_id) lei
from (
select user_id,case client_id when 0 then 'GroupBuy' else name end source,count(*) cnt    
from order_info oi left join client c on oi.client_id=c.id
where date>'2025-10-15' and oi.status='completed' 
    and product_name in('C++','Java','Python')
group by user_id,oi.client_id 
    ) hh
    ) ww
    where lei>=2
    group by source
     order by source