1412.(Hard)查找成绩处于中游的学生

表: Student

+---------------------+---------+
| Column Name         | Type    |
+---------------------+---------+
| student_id          | int     |
| student_name        | varchar |
+---------------------+---------+
student_id 是该表主键.
student_name 学生名字.

表: Exam

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| exam_id       | int     |
| student_id    | int     |
| score         | int     |
+---------------+---------+
(exam_id, student_id) 是该表主键.
学生 student_id 在测验 exam_id 中得分为 score.

成绩处于中游的学生是指至少参加了一次测验, 且得分既不是最高分也不是最低分的学生。

写一个 SQL 语句，找出在 所有 测验中都处于中游的学生 (student_id, student_name)。

不要返回从来没有参加过测验的学生。返回结果表按照 student_id 排序。

查询结果格式如下。

Student 表：
+-------------+---------------+
| student_id  | student_name  |
+-------------+---------------+
| 1           | Daniel        |
| 2           | Jade          |
| 3           | Stella        |
| 4           | Jonathan      |
| 5           | Will          |
+-------------+---------------+

Exam 表：
+------------+--------------+-----------+
| exam_id    | student_id   | score     |
+------------+--------------+-----------+
| 10         |     1        |    70     |
| 10         |     2        |    80     |
| 10         |     3        |    90     |
| 20         |     1        |    80     |
| 30         |     1        |    70     |
| 30         |     3        |    80     |
| 30         |     4        |    90     |
| 40         |     1        |    60     |
| 40         |     2        |    70     |
| 40         |     4        |    80     |
+------------+--------------+-----------+

Result 表：
+-------------+---------------+
| student_id  | student_name  |
+-------------+---------------+
| 2           | Jade          |
+-------------+---------------+

对于测验 1: 学生 1 和 3 分别获得了最低分和最高分。
对于测验 2: 学生 1 既获得了最高分, 也获得了最低分。
对于测验 3 和 4: 学生 1 和 4 分别获得了最低分和最高分。
学生 2 和 5 没有在任一场测验中获得了最高分或者最低分。
因为学生 5 从来没有参加过任何测验, 所以他被排除于结果表。
由此, 我们仅仅返回学生 2 的信息。

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/find-the-quiet-students-in-all-exams

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

Solution

重点在于用rank()窗口函数算出正数排名和倒数排名，再选出min都不为1的学生（既没有正数第一也没有倒数第一过）

with t as 
(SELECT *, 
    RANK() over(PARTITION by exam_id ORDER BY score desc) tier, 
    RANK() over(PARTITION by exam_id ORDER BY score) tier_r
FROM exam)
SELECT s.student_id, max(student_name) student_name
FROM t,Student s where t.student_id=s.student_id
GROUP BY s.student_id
HAVING min(tier) != 1 and min(tier_r) != 1
ORDER BY s.student_id

Table Schema

Create table If Not Exists Student (student_id int, student_name varchar(30));
Create table If Not Exists Exam (exam_id int, student_id int, score int);
Truncate table Student;
insert into Student (student_id, student_name) values ('1', 'Daniel');
insert into Student (student_id, student_name) values ('2', 'Jade');
insert into Student (student_id, student_name) values ('3', 'Stella');
insert into Student (student_id, student_name) values ('4', 'Jonathan');
insert into Student (student_id, student_name) values ('5', 'Will');
Truncate table Exam;
insert into Exam (exam_id, student_id, score) values ('10', '1', '70');
insert into Exam (exam_id, student_id, score) values ('10', '2', '80');
insert into Exam (exam_id, student_id, score) values ('10', '3', '90');
insert into Exam (exam_id, student_id, score) values ('20', '1', '80');
insert into Exam (exam_id, student_id, score) values ('30', '1', '70');
insert into Exam (exam_id, student_id, score) values ('30', '3', '80');
insert into Exam (exam_id, student_id, score) values ('30', '4', '90');
insert into Exam (exam_id, student_id, score) values ('40', '1', '60');
insert into Exam (exam_id, student_id, score) values ('40', '2', '70');
insert into Exam (exam_id, student_id, score) values ('40', '4', '80');