README.md

comments	difficulty	edit_url	tags
true	中等	https://github.com/doocs/leetcode/edit/main/solution/1100-1199/1126.Active%20Businesses/README.md	数据库

1126. 查询活跃业务 🔒

English Version

题目描述

事件表：Events

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| business_id   | int     |
| event_type    | varchar |
| occurrences   | int     | 
+---------------+---------+
(business_id, event_type) 是这个表的主键（具有唯一值的列的组合）。
表中的每一行记录了某种类型的事件在某些业务中多次发生的信息。

 

平均活动 是指有特定 event_type 的具有该事件的所有公司的 occurrences 的均值。

活跃业务 是指具有 多个 event_type 的业务，它们的 occurrences 严格大于 该事件的平均活动次数。

写一个解决方案，找到所有 活跃业务。

以 任意顺序 返回结果表。

结果格式如下所示。

 

示例 1：

输入：
Events table:
+-------------+------------+-------------+
| business_id | event_type | occurrences |
+-------------+------------+-------------+
| 1           | reviews    | 7           |
| 3           | reviews    | 3           |
| 1           | ads        | 11          |
| 2           | ads        | 7           |
| 3           | ads        | 6           |
| 1           | page views | 3           |
| 2           | page views | 12          |
+-------------+------------+-------------+
输出：
+-------------+
| business_id |
+-------------+
| 1           |
+-------------+ 
解释：
每次活动的平均活动可计算如下:
- 'reviews': (7+3)/2 = 5
- 'ads': (11+7+6)/3 = 8
- 'page views': (3+12)/2 = 7.5
id=1 的业务有 7 个 'reviews' 事件(多于 5 个)和 11 个 'ads' 事件(多于 8 个)，所以它是一个活跃的业务。

解法

方法一

MySQL

# Write your MySQL query statement below
SELECT business_id
FROM
    EVENTS AS t1
    JOIN (
        SELECT
            event_type,
            AVG(occurences) AS occurences
        FROM EVENTS
        GROUP BY event_type
    ) AS t2
        ON t1.event_type = t2.event_type
WHERE t1.occurences > t2.occurences
GROUP BY business_id
HAVING COUNT(1) > 1;

方法二

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            business_id,
            occurences > AVG(occurences) OVER (PARTITION BY event_type) AS mark
        FROM Events
    )
SELECT business_id
FROM T
WHERE mark = 1
GROUP BY 1
HAVING COUNT(1) > 1;