查找至少参加过一项考试但未参加最高或最低分数的学生

Question

我的解决方案通过了第一个测试用例，但在最终提交后得到了错误的答案。我感谢任何愿意指出我的错误的人。谢谢！

问题如下：

表：学生

+---------------------+---------+
| Column Name         | Type    |
+---------------------+---------+
| student_id          | int     |
| student_name        | varchar |
+---------------------+---------+

student_id 是该表的主键。 student_name 是学生的姓名。

表格：考试

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| exam_id       | int     |
| student_id    | int     |
| score         | int     |
+---------------+---------+

(exam_id, student_id) 是该表的主键。 拥有 student_id 的学生在 id 为exam_id 的考试中获得分数。

“相当”的学生是指至少参加过一次考试并且既没有获得高分也没有获得低分的学生。

编写 SQL 查询以报告学生（student_id、student_name）在所有考试中“安静”。

不要让从未参加过任何考试的学生退学。返回按 student_id 排序的结果表。

查询结果格式如下例。

学生桌：

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

考试表：

+------------+--------------+-----------+
| 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     |
+------------+--------------+-----------+

结果表：

+-------------+---------------+
| student_id  | student_name  |
+-------------+---------------+
| 2           | Jade          |
+-------------+---------------+

问题中的解释：

对于考试 1：学生 1 和 3 分别获得最低和最高分。 对于考试 2：学生 1 持有最高分和最低分。 对于考试 3 和 4：Studnet 1 和 4 分别获得最低和最高分。 学生 2 和 5 从未在任何考试中获得最高或最低的分数。 由于学生 5 没有参加任何考试，因此他被排除在结果之外。 所以，我们只返回学生 2 的信息。

我的答案是创建两张表，一张是列出符合条件的学生，至少有一次考试。 另一种是找到考试表的最大值（分数）和最小值（分数）。 并使用 定位安静学生的 id，然后加入 Student 表查找 这个 student_id 的名字，如下：

-- get eligible student

with eligible_student as
(
select distinct student_id as eligible_id from Exam
    group by 1
    order by 1
),

-- get the high and low score
high_low as
(select student_id, max(score) as high_score, min(score) as low_score
from Exam), 

result as 
 (select eligible_student.eligible_id as student_id
 from eligible_student inner join
 high_low 
 on eligible_student.eligible_id <> high_low.student_id
-- left join Student
-- on eligible_student.eligible_id = Student.student_id 
 group by student_id
 order by student_id
 )
 
 select result.student_id, s.student_name as student_name
 from result left join Student s
 on result.student_id = s.student_id
 order by student_id;

Answer 1

我会使用窗口函数和聚合：

select s.*
from student s
inner join (
    select e.*, 
        rank() over(partition by exam_id order by score) as rn_asc,
        rank() over(partition by exam_id order by score desc) as rn_desc
    from exam e
) e on e.student_id = s.student_id
group by s.student_id
having min(rn_asc) > 1 and min(rn_desc) > 1

子查询通过升序和降序对具有相同考试的记录进行排名，允许平局。然后，我们可以将其与学生表连接起来（这会消除根本没有考试的学生），按学生分组，并过滤​​出两个排名都未达到 1 的学生。

Answer 2

这个查询：

SELECT *,
     (MIN(score) OVER (PARTITION BY exam_id) = score) +
     (MAX(score) OVER (PARTITION BY exam_id) = score) flag
FROM exam

当学生的分数既不是考试的最低分数也不是最高分数时，返回一个flag 列，其值为0。

您可以对上述查询的结果进行聚合，以获取所有没有一个flag 且值不同于0 的学生：

WITH cte AS (
  SELECT *,
         (MIN(score) OVER (PARTITION BY exam_id) = score) +
         (MAX(score) OVER (PARTITION BY exam_id) = score) flag
  FROM exam
)
SELECT s.student_id, s.student_name
FROM student s INNER JOIN cte c
ON c.student_id = s.student_id
GROUP BY s.student_id, s.student_name
HAVING SUM(c.flag) = 0

或者：

WITH cte AS (
  SELECT student_id
  FROM (
    SELECT *,
         (MIN(score) OVER (PARTITION BY exam_id) = score) +
         (MAX(score) OVER (PARTITION BY exam_id) = score) flag
    FROM exam
  ) t
  GROUP BY student_id
  HAVING SUM(flag) = 0
)
SELECT * 
FROM student
WHERE student_id IN (SELECT student_id FROM cte)

请参阅demo。
结果：

> student_id | student_name
> ---------: | :-----------
>          2 | Jade        

Answer 3

这部分错了：

with high_low as
(select student_id, max(score) as high_score, min(score) as low_score
from Exam)

因为它输出：

+------------+------------+-----------+
| student_id | high_score | low_score |
+------------+------------+-----------+
|          1 |         90 |        60 |
+------------+------------+-----------+

并且 student_id=1 与找到的 high_score 或 low_score 无关。

在此之后，找到的（但不正确的）student_id 用于选择 cte result。

解决方案：

with high_low as
(select max(score) as high_score, min(score) as low_score
from Exam) 
select student.student_id, student_name
from (
  select exam.student_id 
   from exam
   group by exam.student_id
   having max(score) <> (select high_score from high_low)
      and min(score) <> (select low_score from high_low)) x
inner join student on student.student_id=x.student_id;

或：

select 
   student.*
from exam 
inner join student on student.student_id=exam.student_id
group by student_id 
having max(score) not in (select max(score) from exam) 
   and min(score) not in (select min(score) from exam);