题解 | #牛客的课程订单分析(四)#

select * 
from (
		select A.user_id,min(o.date) first_buy_date,A.nn cnt
		from order_info  o
		left join(
				select  user_id,count(user_id) nn from  order_info
				where date > '2025-10-15' and 
				status = 'completed'  and product_name in('C++','Python','Java')
				group by user_id
				having count(user_id)>1
				order by user_id
		) A on o.user_id  = A.user_id
		where o.date > '2025-10-15' and 
				o.status = 'completed'  and o.product_name in('C++','Python','Java')
		group by A.user_id
		order by A.user_id) B
where B.user_id is not null;