本文主要是介绍【牛客】SQL136 每类试卷得分前3名-窗口函数，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
描述
现有试卷信息表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 | 2021-09-01 06:00:00 |
| 2 | 9002 | SQL | hard | 60 | 2021-09-01 06:00:00 |
| 3 | 9003 | 算法 | medium | 80 | 2021-09-01 10:00:00 |
试卷作答记录表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 | 2021-09-01 09:01:01 | 2021-09-01 09:31:00 | 78 |
| 2 | 1002 | 9001 | 2021-09-01 09:01:01 | 2021-09-01 09:31:00 | 81 |
| 3 | 1002 | 9002 | 2021-09-01 12:01:01 | 2021-09-01 12:31:01 | 81 |
| 4 | 1003 | 9001 | 2021-09-01 19:01:01 | 2021-09-01 19:40:01 | 86 |
| 5 | 1003 | 9002 | 2021-09-01 12:01:01 | 2021-09-01 12:31:51 | 89 |
| 6 | 1004 | 9001 | 2021-09-01 19:01:01 | 2021-09-01 19:30:01 | 85 |
| 7 | 1005 | 9003 | 2021-09-01 12:01:01 | 2021-09-01 12:31:02 | 85 |
| 8 | 1006 | 9003 | 2021-09-07 10:01:01 | 2021-09-07 10:21:01 | 84 |
| 9 | 1003 | 9003 | 2021-09-08 12:01:01 | 2021-09-08 12:11:01 | 40 |
| 10 | 1003 | 9002 | 2021-09-01 14:01:01 | (NULL) | (NULL) |
找到每类试卷得分的前3名，如果两人最大分数相同，选择最小分数大者，如果还相同，选择uid大者。由示例数据结果输出如下：
| tid | uid | ranking |
| SQL | 1003 | 1 |
| SQL | 1004 | 2 |
| SQL | 1002 | 3 |
| 算法 | 1005 | 1 |
| 算法 | 1006 | 2 |
| 算法 | 1003 | 3 |
解释：有作答得分记录的试卷tag有SQL和算法，SQL试卷用户1001、1002、1003、1004有作答得分，最高得分分别为81、81、89、85，最低得分分别为78、81、86、40，因此先按最高得分排名再按最低得分排名取前三为1003、1004、1002。
方法一：
select * from(selecttid,uid,row_number() over(partition by tid order by down_rnk) as rankingfrom(selecttid,uid,dense_rank() over(partition by tid order by max_score desc) as down_rnk,dense_rank() over(partition by tid order by min_score desc) as up_rnkfrom(selecttag as tid,uid,max(score) as max_score,min(score) as min_scorefromexam_record left join examination_info using(exam_id)where submit_time is not nullgroup by uid,tid)t1order by tid,down_rnk,up_rnk,uid desc)t2)t3
where ranking<=3方法二：
select *
from(select tag as tid,uid,row_number() over(partition by tag order by max(score) desc,min(score) desc,uid desc) as rankingfromexam_record left join examination_info using(exam_id)where submit_time is not nullgroup by uid,tid)t1
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