各专业平均成绩较高的学生

如何解决各专业平均成绩较高的学生

表和数据：

CREATE TABLE major
( 
    id INT PRIMARY KEY,name VARCHAR(200)
);

insert into major values 
    (101,'Computing'),(102,'Arquitecture');

CREATE TABLE student
( 
    id INT PRIMARY KEY,name VARCHAR(200),major_id INT,foreign key(major_id) references major(id)
);

insert into student  values
    (1001,'Claude',101),(1002,'John',(1003,'Peter',102);

CREATE TABLE course
( 
    id INT PRIMARY KEY,name VARCHAR(200)
);

insert into course values 
    (901,'Databases'),(902,'Java'),(903,'Artificial Intelligence'),(904,'OOP');
    
CREATE TABLE grades
(
   student_id INT,course_id INT,grade integer,primary key (student_id,course_id),foreign key(student_id) references student(id),foreign key(course_id) references course(id)
);

insert into grades  values
    (1001,903,95),(1001,904,88),901,76),82),902,87);

预期：

| student    | major        | grade |
| ---------- | -------------| ----- |
| Peter      | Architecture | 87    |
| Claude     | Computing    | 91.5  |

换句话说：检索每个专业的高年级学生。

游乐场here。

如果可能，没有TOP，LIMIT。

如果可能的话，请使用旧的ANSI SQL以及使用窗口函数。

使用MySQL引擎，但不是必需的。

我的方法#1：

-- average grade by student
select s.name as Student,m.name as Major,avg(g.grade) as Average
         from student s
         inner join grades g on (s.id = g.student_id)
         inner join major m on (m.id = s.major_id)
         group by s.id

但不需要约翰：

| Student | Major        | Average |
| ------- | ------------ | ------- |
| Claude  | Computing    | 91.5000 |
| John    | Computing    | 79.0000 |
| Peter   | Arquitecture | 87.0000 |

我的方法#2：

-- Max average grade by career; lacks student
select a.major,max (a.average) as Average
    from (select s.name as Student,avg(g.grade) as average
         from student s
         inner join grades g on (s.id = g.student_id)
         inner join major m on (m.id = s.major_id)
         group by s.id) a
    group by a.major;           

但缺少学生专栏。

| major        | Average |
| ------------ | ------- |
| Arquitecture | 87.0000 |
| Computing    | 91.5000 |

谢谢。

解决方法

如果您正在运行MySQL 8.0，则可以使用rank()进行此操作：

select *
from (
    select s.name as student,m.name as major,avg(g.grade) as average,rank() over(partition by m.id order by avg(g.grade) desc) rn
    from student s
    inner join grades g on s.id = g.student_id
    inner join major m on  m.id = s.major_id
    group by s.id,m.id
) t
where rn = 1

注意：

• rank()允许联系

联接条件周围的

• 括号是多余的

,

您可以结合使用两个查询：

    [0,5]      (5,10]      (10,15]
a     0          1            0
b     0          0            0
c     1          2            2
d     0          1            0

,

对于旧版本的MySQL ...

SELECT x.* 
  FROM
     ( SELECT m.name major,s.name student,AVG(grade) avg_grade
         FROM major m
         JOIN student s
           ON s.major_id = m.id
         JOIN grades g
           ON g.student_id = s.id
        GROUP
           BY major,student
     ) x
  JOIN
     (
       SELECT major,MAX(avg_grade) avg_grade
         FROM 
            ( SELECT m.name major,AVG(grade) avg_grade
                FROM major m
                JOIN student s
                  ON s.major_id = m.id
                JOIN grades g
                  ON g.student_id = s.id
               GROUP
                  BY major,student
            )n
        GROUP
           BY major
      ) y
     ON y.major = x.major
    AND y.avg_grade = x.avg_grade

| major        | student | avg_grade |
| ------------ | ------- | --------- |
| Arquitecture | Peter   | 87.0000   |
| Computing    | Claude  | 91.5000   |

---

View on DB Fiddle