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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 rnk2from 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 rnk2from 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