构建二维MySQL查询的最有效方法是什么？

I have a MySQL database with the following tables and fields:

我有一个MySQL数据库,包含以下表格和字段:

Student (id)
Class (id)
Grade (id, student_id, class_id, grade)

成绩(id,student_id,class_id,年级)

The student and class tables are indexed on id (primary keys). The grade table is indexed on id (primary key) and student_id, class_id and grade.

学生和班级表在id(主键)上编制索引。成绩表是在id(主键)和student_id,class_id和grade上编制索引的。

I need to construct a query which, given a class ID, gives a list of all other classes and the number of students who scored more in that other class.

我需要构建一个查询,给定一个类ID,给出所有其他类的列表以及在其他类中获得更多分数的学生数。

Essentially, given the following data in the grades table:

基本上,鉴于成绩表中的以下数据:

student_id | class_id | grade
--------------------------------------
1          | 1        | 87
1          | 2        | 91
1          | 3        | 75
2          | 1        | 68
2          | 2        | 95
2          | 3        | 84
3          | 1        | 76
3          | 2        | 88
3          | 3        | 71

Querying with class ID 1 should yield:

使用类ID 1查询应该产生:

class_id | total
-------------------
2        | 3
3        | 1

Ideally I'd like this to execute in a few seconds, as I'd like it to be part of a web interface.

理想情况下,我希望在几秒钟内执行此操作,因为我希望它能够成为Web界面的一部分。

The issue I have is that in my database, I have over 1300 classes and 160,000 students. My grade table has almost 15 million rows and as such, the query takes a long time to execute.

我的问题是,在我的数据库中,我有超过1300个班级和160,000名学生。我的成绩表有近1500万行,因此查询需要很长时间才能执行。

Here's what I've tried so far along with the times each query took:

这是我到目前为止所尝试的以及每个查询所花费的时间:

-- I manually stopped execution after 2 hours
SELECT    c.id, COUNT(*) AS total
FROM      classes c
              INNER JOIN grades a ON a.class_id = c.id
              INNER JOIN grades b ON b.grade < a.grade AND
                  a.student_id = b.student_id AND
                  b.class_id = 1
WHERE     c.id != 1 AND
GROUP BY  c.id

-- I manually stopped execution after 20 minutes
SELECT    c.id,
          (
              SELECT    COUNT(*) 
              FROM      grades g 
              WHERE     g.class_id = c.id AND g.grade > (
                            SELECT   grade 
                            FROM     grades 
                            WHERE    student_id = g.student_id AND 
                                     class_id = 1
                        )
          ) AS total
FROM      classes c
WHERE     c.id != 1;

-- 1 min 12 sec
CREATE TEMPORARY TABLE temp_blah (student_id INT(11) PRIMARY KEY, grade INT);
INSERT INTO temp_blah SELECT student_id, grade FROM grades WHERE class_id = 1;
SELECT    o.id,
          ( 
              SELECT    COUNT(*)
              FROM      grades g
                            INNER JOIN temp_blah t ON g.student_id = t.student_id
              WHERE     g.class_id = c.id AND t.grade < g.grade
          ) AS total
FROM      classes c
WHERE     c.id != 1;

-- Same thing but with joins instead of a subquery - 1 min 54 sec
SELECT    c.id,
          COUNT(*) AS total
FROM      classes c
              INNER JOIN grades g ON c.id = p.class_id
              INNER JOIN temp_blah t ON g.student_id = t.student_id
WHERE     c.id != 1
GROUP BY  c.id;

I also considered creating a 2D table, with students as rows and classes as columns, however I can see two issues with this:

我还考虑创建一个2D表,学生作为行和类作为列,但我可以看到两个问题:

MySQL implements a maximum column count (4096) and maximum row size (in bytes) which may be exceeded by this approach

MySQL实现了这种方法可能超过的最大列数(4096)和最大行大小(以字节为单位)

I can't think of a good way to query that structure to get the results I need

我想不出一个查询该结构以获得我需要的结果的好方法

I also considered performing these calculations as background jobs and storing the results somewhere, but for the information to remain current (it must), they would need to be recalculated every time a student, class or grade record was created or updated.

我还考虑将这些计算作为后台作业并将结果存储在某处,但为了使信息保持最新(必须),每次创建或更新学生,班级或成绩记录时都需要重新计算。

Does anyone know a more efficient way to construct this query?

有谁知道构建此查询的更有效方法?

EDIT: Create table statements:

编辑:创建表语句:

CREATE TABLE `classes` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=1331 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci$$

CREATE TABLE `students` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=160803 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci$$

CREATE TABLE `grades` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `student_id` int(11) DEFAULT NULL,
  `class_id` int(11) DEFAULT NULL,
  `grade` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `index_grades_on_student_id` (`student_id`),
  KEY `index_grades_on_class_id` (`class_id`),
  KEY `index_grades_on_grade` (`grade`)
) ENGINE=InnoDB AUTO_INCREMENT=15507698 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci$$

Output of explain on the most efficient query (the 1 min 12 sec one):

关于最有效查询的解释输出(1分12秒一):

id | select_type        | table | type   | possible_keys                                                             | key                      | key_len | ref               | rows   | extra 
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1  | PRIMARY            | c     | range  | PRIMARY                                                                   | PRIMARY                  | 4       |                   | 683    | Using where; Using index
2  | DEPENDENT SUBQUERY | g     | ref    | index_grades_on_student_id,index_grades_on_class_id,index_grades_on_grade | index_grades_on_class_id | 5       | mydb.c.id         | 830393 | Using where
2  | DEPENDENT SUBQUERY | t     | eq_ref | PRIMARY                                                                   | PRIMARY                  | 4       | mydb.g.student_id | 1      | Using where

Another edit - explain output for sgeddes suggestion:

另一个编辑 - 解释sgeddes建议的输出:

+----+-------------+------------+--------+---------------+------+---------+------+----------+----------------------------------------------+
| id | select_type | table      | type   | possible_keys | key  | key_len | ref  | rows     | Extra                                        |
+----+-------------+------------+--------+---------------+------+---------+------+----------+----------------------------------------------+
|  1 | PRIMARY     | <derived2> | ALL    | NULL          | NULL | NULL    | NULL | 14953992 | Using where; Using temporary; Using filesort |
|  2 | DERIVED     | <derived3> | system | NULL          | NULL | NULL    | NULL |        1 | Using filesort                               |
|  2 | DERIVED     | G          | ALL    | NULL          | NULL | NULL    | NULL | 15115388 |                                              |
|  3 | DERIVED     | NULL       | NULL   | NULL          | NULL | NULL    | NULL |     NULL | No tables used                               |
+----+-------------+------------+--------+---------------+------+---------+------+----------+----------------------------------------------+

2 个解决方案

#1

I think this should work for you using SUM and CASE:

我认为这应该适用于你使用SUM和CASE:

SELECT C.Id,
  SUM(
    CASE 
    WHEN G.Grade > C2.Grade THEN 1 ELSE 0 
    END
  ) 
FROM Class C
  INNER JOIN Grade G ON C.Id = G.Class_Id
  LEFT JOIN (
      SELECT Grade, Student_Id, Class_Id
      FROM Class
        JOIN Grade ON Class.Id = Grade.Class_Id
      WHERE Class.Id = 1
    ) C2 ON G.Student_Id = C2.Student_Id
WHERE C.Id <> 1
GROUP BY C.Id

Sample Fiddle Demo

示例小提琴演示

--EDIT--

In response to your comment, here is another attempt that should be much faster:

在回应您的评论时,这是另一个应该更快的尝试:

SELECT 
  Class_Id, 
  SUM(CASE WHEN Grade > minGrade THEN 1 ELSE 0 END)
FROM 
(
  SELECT 
    Student_Id,
    @classToCheck:=
      IF(G.Class_Id = 1, Grade, @classToCheck) minGrade ,
    Class_Id,
    Grade
  FROM Grade G
    JOIN (SELECT @classToCheck:= 0) t
  ORDER BY Student_Id, IF(Class_Id = 1, 0, 1)
  ) t
WHERE  Class_Id <> 1
GROUP BY Class_ID

And more sample fiddle.

还有更多样品小提琴。

#2

Can you give this a try on the original data as well! It is only one join :)

你能不能试试原始数据吧!这只是一个加入:)

select
  final.class_id, count(*) as total
from
  (
    select * from   
      (select student_id as p_student_id, grade as p_grade from table1 where class_id = 1) as partial
    inner join table1 on table1.student_id = partial.p_student_id
    where table1.class_id <> 1 and table1.grade > partial.p_grade    
  ) as final
 group by
  final.class_id;

sqlfiddle link

#1