【牛客】SQL123 SQL类别高难度试卷得分的截断平均值

描述

牛客的运营同学想要查看大家在SQL类别中高难度试卷的得分情况。

请你帮她从exam_record数据表中计算所有用户完成SQL类别高难度试卷得分的截断平均值（去掉一个最大值和一个最小值后的平均值）。

示例数据：examination_info（exam_id试卷ID, tag试卷类别, difficulty试卷难度, duration考试时长, release_time发布时间）

id	exam_id	tag	difficulty	duration	release_time
1	9001	SQL	hard	60	2020-01-01 10:00:00
2	9002	算法	medium	80	2020-08-02 10:00:00

示例数据：exam_record（uid用户ID, exam_id试卷ID, start_time开始作答时间, submit_time交卷时间, score得分）

id	uid	exam_id	start_time	submit_time	score
1	1001	9001	2020-01-02 09:01:01	2020-01-02 09:21:01	80
2	1001	9001	2021-05-02 10:01:01	2021-05-02 10:30:01	81
3	1001	9001	2021-06-02 19:01:01	2021-06-02 19:31:01	84
4	1001	9002	2021-09-05 19:01:01	2021-09-05 19:40:01	89
5	1001	9001	2021-09-02 12:01:01	(NULL)	(NULL)
6	1001	9002	2021-09-01 12:01:01	(NULL)	(NULL)
7	1002	9002	2021-02-02 19:01:01	2021-02-02 19:30:01	87
8	1002	9001	2021-05-05 18:01:01	2021-05-05 18:59:02	90
9	1003	9001	2021-09-07 12:01:01	2021-09-07 10:31:01	50
10	1004	9001	2021-09-06 10:01:01	(NULL)	(NULL)

根据输入你的查询结果如下：

tag	difficulty	clip_avg_score
SQL	hard	81.7

从examination_info表可知，试卷9001为高难度SQL试卷，该试卷被作答的得分有[80,81,84,90,50]，去除最高分和最低分后为[80,81,84]，平均分为81.6666667，保留一位小数后为81.7

输入描述：

输入数据中至少有3个有效分数

with cte as (
    select exam_id,score,tag,difficulty,
    dense_rank() over (partition by exam_id order by score) as rnk1,
    dense_rank() over (partition by exam_id order by score desc) as rnk2
    from 
    exam_record left join examination_info using (exam_id)
    where score is not null and tag = 'SQL'and difficulty = 'hard'
)

select tag,difficulty,
round(avg(score),1) as clip_avg_score
from cte
where score between 
(select score from cte where rnk1=2) 
and (select score from cte where rnk2=2)
group by tag,difficulty

或：

with cte as (
    select exam_id,score,tag,difficulty,
    dense_rank() over (partition by exam_id order by score) as rnk1,
    dense_rank() over (partition by exam_id order by score desc) as rnk2
    from 
    exam_record left join examination_info using (exam_id)
    where score is not null and tag = 'SQL'and difficulty = 'hard'
)

select tag,difficulty,
round(avg(score),1) as clip_avg_score
from cte
where rnk1!=1 and rnk2!=1
group by tag,difficulty