结合两个简单的SQL查询两个一对一的查询

Question

我有两个查询，我想将它们合并为一个查询

第一个查询：此查询查找每个科目的每个学生的平均值：

SELECT  
    StudentFirstName, StudentLastName, ClassName, 
    AVG(Grade) AS 'average for this subject'
FROM tests
INNER JOIN students ON tests.StudentID = students.StudentID
GROUP BY StudentFirstName, StudentLastName, ClassName;

第二个查询：此查询查找每个学生的平均分数，以得出其最终总成绩的总和：

SELECT  
    StudentFirstName, StudentLastName, AVG(average) AS total_average
FROM
    (SELECT   
         StudentFirstName, StudentLastName, AVG(Grade) AS average
     FROM 
         tests
     INNER JOIN 
         students ON tests.StudentID = students.StudentID
     GROUP BY 
         StudentFirstName, StudentLastName, ClassName) AS t
GROUP BY 
    StudentFirstName, StudentLastName;

例如：我的成绩（即Error404）为：

• 代数：第一次考试：99，第二次考试：97，第三次考试：96 ---> 第一个查询给出的平均值： 97.3333

• 机器学习：第一次考试：95，第二次考试：94 ---> 第一个查询给出的平均分数： 94.5

第二个查询返回学生名为error404的AVG的总平均值（97.3333,94.5）= 95.91665

表1- 学生 ：

      pk-INT           VARCHAR             VARCHAR
    +-----------+------------------+-----------------+
    | StudentID | StudentFirstName | StudentLastName |
    +-----------+------------------+-----------------+
    |         1 | agam             | rafaeli         |
    |         2 | amir             | aizinger        |
    |         3 | avi              | caspi           |
    |         4 | avia             | wolf            |
    +-----------+------------------+-----------------+

表2- 测试 ：

 PK-VARCHR    PK-VARCHR     PK&FK-INT   INT
+------------+------------+-----------+-------+
| TestDate   | ClassName  | StudentID | Grade |
+------------+------------+-----------+-------+
| 2017-07-01 | Algebra    |         1 |    88 |
| 2017-08-02 | Algo       |         1 |    97 |
| 2017-09-01 | Algebra    |         1 |    80 |
| 2017-09-01 | Algebra    |         1 |    97 |
| 2017-09-01 | Set-theory |         1 |    85 |
| 2017-09-04 | Calcules   |         1 |    86 |
| 2016-05-03 | Set-theory |         2 |    84 |
| 2016-07-02 | Calcules   |         2 |    89 |
| 2016-07-04 | Algo       |         2 |    83 |
| 2016-07-05 | Algebra    |         2 |    79 |
| 2016-06-03 | Algebra    |         3 |    99 |
| 2016-07-02 | Algo       |         3 |    97 |
| 2016-07-03 | Calcules   |         3 |    96 |
| 2016-09-03 | Set-theory |         3 |    95 |
| 2016-06-03 | Algebra    |         4 |    78 |
+------------+------------+-----------+-------+

样本数据：

DROP DATABASE IF EXISTS error404;
CREATE DATABASE error404;

USE error404

CREATE TABLE students
(
    StudentID        INT NOT NULL AUTO_INCREMENT,
    StudentFirstName VARCHAR(25),
    StudentLastName  VARCHAR(25),
    PRIMARY KEY (StudentID)
);

INSERT INTO students (StudentFirstName, StudentLastName) 
VALUES ('agam', 'rafaeli'), ('amir', 'aizinger'), ('avi', 'caspi'),
       ('avia', 'wolf ');

CREATE TABLE tests
(
    testid    INT NOT NULL AUTO_INCREMENT,
    TestDate  DATE,
    ClassName VARCHAR(25),
    StudentID INT NOT NULL,
    Grade     INT NOT NULL,
    PRIMARY KEY (testid),
    KEY (StudentID)
);

INSERT INTO tests (TestDate, ClassName, StudentID, Grade) 
VALUES ('2017-07-01', 'Algebra', 1, 88), 
       ('2017-08-02', 'Algo', 1, 97),
       ('2017-09-01', 'Algebra', 1, 80), 
       ('2017-09-01', 'Algebra', 1, 97),
       ('2017-09-01', 'Set-theory', 1, 85), 
       ('2017-09-04', 'Calculus', 1, 86),
       ('2016-05-03', 'Set-theory', 2, 84), 
       ('2016-07-02', 'Calculus', 2, 89),
       ('2016-07-04', 'Algo', 2, 83), 
       ('2016-07-05', 'Algebra', 2, 79),
       ('2016-06-03', 'Algebra', 3, 99), 
       ('2016-07-02', 'Algo', 3, 97),
       ('2016-07-03', 'Calculus', 3, 96), 
       ('2016-09-03', 'Set-theory', 3, 95),
       ('2016-06-03', 'Algebra', 4, 78);

我想要这个结果：

+--------------+--------------+--------------+--------------+--------------+
| StFirstName  | StLastName   | ClassName    | aveInSubject | totalAve     |
+--------------+--------------+--------------+--------------+--------------+
| name1        | lname1       | algebra      |  80          |  87          |
| name1        | lname1       | algo         |  88          |  87          |
| name1        | lname1       | calcul       |  93          |  87          |
| name2        | lname2       | algebra      |  70          |  74.3        |
| name2        | lname2       | algo         |  76          |  74.3        |
| name2        | lname2       | calcul       |  77          |  74.3        |
+--------------+--------------+--------------+--------------+--------------+

Answer 1

您可以join每科平均每个学生平均总的结果。

SELECT t1.*,t2.totalAvg
FROM (SELECT StudentFirstName,StudentLastName,ClassName,AVG(Grade) AS `average for this subject`
      FROM tests
      INNER JOIN students ON tests.StudentID=students.StudentID
      GROUP BY StudentFirstName,StudentLastName,ClassName
     ) t1
JOIN (SELECT StudentFirstName,StudentLastName,AVG(`average for this subject`) as totalAvg
      FROM (SELECT StudentFirstName,StudentLastName,ClassName,AVG(Grade) AS `average for this subject`
            FROM tests
            INNER JOIN students ON tests.StudentID=students.StudentID
            GROUP BY StudentFirstName,StudentLastName,ClassName
           ) t
      GROUP BY StudentFirstName,StudentLastName
     ) t2 
ON t1.StudentFirstName=t2.StudentFirstName and t1.StudentLastName=t2.StudentLastName

编辑：对于具有窗口功能的MySQL的将来版本（从8.0版开始），查询可以简化为

select studentfirstname,studentlastname,classname,avgPerSubject
,sum(avgPerSubject) over w/count(*) over w as totalAvg
from (select distinct 
      s.studentfirstname,s.studentlastname,t.classname,
      avg(t.grade) over(partition by s.studentfirstname,s.studentlastname,t.classname) as avgPerSubject
      from tests t
      join students s on s.studentid=t.studentid
     ) t
window w as (partition by studentfirstname,studentlastname)