单挑力扣（LeetCode）SQL题：1126. 查询活跃业务（难度：中等）

事件表：Events
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| business_id   | int     |
| event_type    | varchar |
| occurences    | int     | 
+---------------+---------+
此表的主键是 (business_id, event_type)。
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。


写一段 SQL 来查询所有活跃的业务。
如果一个业务的某个事件类型的发生次数大于此事件类型在所有业务中的平均发生次数，并且该业务至少有两个这样的事件类型，那么该业务就可被看做是活跃业务。


查询结果格式如下所示：
Events table:
+-------------+------------+------------+
| business_id | event_type | occurences |
+-------------+------------+------------+
| 1           | reviews    | 7          |
| 3           | reviews    | 3          |
| 1           | ads        | 11         |
| 2           | ads        | 7          |
| 3           | ads        | 6          |
| 1           | page views | 3          |
| 2           | page views | 12         |
+-------------+------------+------------+


结果表
+-------------+
| business_id |
+-------------+
| 1           |
+-------------+ 
'reviews'、 'ads' 和 'page views' 的总平均发生次数分别是 (7+3)/2=5, (11+7+6)/3=8, (3+12)/2=7.5。
id 为 1 的业务有 7 个 'reviews' 事件（大于 5）和 11 个 'ads' 事件（大于 8），所以它是活跃业务。


来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/active-businesses

#测试数据
Create table If Not Exists Events (business_id int, event_type varchar(10), occurences int);
insert into Events (business_id, event_type, occurences) values ('1', 'reviews', '7');
insert into Events (business_id, event_type, occurences) values ('3', 'reviews', '3');
insert into Events (business_id, event_type, occurences) values ('1', 'ads', '11');
insert into Events (business_id, event_type, occurences) values ('2', 'ads', '7');
insert into Events (business_id, event_type, occurences) values ('3', 'ads', '6');
insert into Events (business_id, event_type, occurences) values ('1', 'page views', '3');
insert into Events (business_id, event_type, occurences) values ('2', 'page views', '12');

#方法一：使用子查询
with 
tmp1 as (
select
        a.event_type,
avg(a.occurences) avg_occurences
from Events a
group by a.event_type
)
select
    b.business_id
from Events b
inner join tmp1 c
on b.event_type = c.event_type
and b.occurences > c.avg_occurences
group by b.business_id
having count(1)>=2;
#方法二：使用分析函数
select b.business_id
from 
(
select
        a.business_id,
        a.occurences,
avg(a.occurences) over(partition by a.event_type) as avg_occurences
from events a
)b
where b.occurences > b.avg_occurences
group by 1
having count(1) > 1;
#方法三：使用自关联
SELECT
    c.business_id
from
(
SELECT
        a.business_id,
        a.event_type
FROM Events a
inner join Events b
on a.event_type = b.event_type
group by a.business_id,a.event_type
having max(a.occurences) > avg(b.occurences)
)c
group by c.business_id
having count(1)>=2;