LeetCode（数据库）- 查询活跃业务

Posted 2021-09-01 程序员牧码

题目链接：点击打开链接

题目大意：略。

解题思路：略。

AC 代码

WITH t1 AS(SELECT DISTINCT event_type, AVG(occurences) OVER(PARTITION BY event_type) avgn FROM Events)

SELECT business_id 
FROM Events e JOIN t1 ON e.event_type = t1.event_type
WHERE occurences > t1.avgn
GROUP BY business_id
HAVING COUNT(*) > 1