题解 | #牛客的课程订单分析(五)#
牛客的课程订单分析(五)
http://www.nowcoder.com/practice/348afda488554ceb922efd2f3effc427
思路: 求second_buy_date的核心重点是:min(date > min(date))
我们使用订单分析四中的结果,求出最小的date,然后联立 order_info 表,通过 t1.date > t2.first_buy_date
及 min(t1.date)
求出第二小date
完整代码:
select t1.user_id, t2.first_buy_date, min(t1.date) as second_buy_date, t2.cnt
from order_info t1, (select user_id, min(date) as first_buy_date, count(user_id) as cnt
from order_info
where status = 'completed'
and date > '2025-10-15'
and product_name in ('C++', 'Java', 'Python')
group by user_id
having count(user_id) > 1) as t2
where t1.user_id = t2.user_id
and t1.date > t2.first_buy_date
and t1.status = 'completed'
and t1.date > '2025-10-15'
and t1.product_name in ('C++', 'Java', 'Python')
group by user_id
order by user_id
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已完成牛客的SQL练习。接下来是算法的练习