题解 | #获取当前薪水第二多的员工的emp_no以及其对应的薪水salary#

select user_id,min(date) as first_buy_date,
max(date) as second_buy_date,cnt
from (select user_id,date,row_number() over (partition by user_id order by date) sort_number,
     count(*) over (partition by user_id) cnt
     from order_info where product_name in ("C++","Python","Java")
     and date > "2025-10-15" and status = "completed") t1
     where t1.cnt>1 and sort_number<3 GROUP BY user_id order by user_id

参考了一下别人的方法，使用窗口函数会简便好多
按照日期排序，取出前两个，小的作为first_buy_date,大的作为second_buy_date