题解 | #牛客的课程订单分析(四)#

SELECT
oi.user_id,
min(oi.date),
x.cnt
FROM order_info oi
LEFT JOIN
(select user_id,
COUNT(id) as cnt
from order_info
where datediff(date,"2025-10-15")>0
  and status ="completed"
  and product_name in("C++","Java","Python")
group by user_id
having count(id)>=2)x ON x.user_id = oi.user_id

where datediff(oi.date,"2025-10-15")>0
  and oi.status ="completed"
  and oi.product_name in("C++","Java","Python")
  and x.cnt is NOT NULL
group by oi.user_id,x.cnt
ORDER BY oi.user_id

这个方式比刚刚更简洁一些