题解 | #牛客的课程订单分析(七)#

select coalesce(name,'GroupBuy') as source,count(1) as cnt 
from 
(    select *
    ,count(1)over(partition by user_id) as ct 
    from order_info
    where date >'2025-10-15'
    and status='completed'
    and product_name in ('C++','Java','Python')
) a left join client on a.client_id=client.id
where ct>=2
group by name
order by source

和上一题的写法差不多，这里直接group by name，就ok了，null会自动分为一组，使用coalesce(a,b,c,.......)来修改别名，

coalesce(name,'GroupBuy') 这里的意思是name列如果有null值，则返回GroupBuy