mysql - SQL选择学生的平均成绩

我正在学习sql，对此有点麻烦，
我需要每个科目的每个学生的平均成绩。

我做了下表。

students
|id_student|name|

subjects
|id_subject|name|

grades
|id_grade|value|

即时通讯使用这些表链接它们：

students_subjects
|id_student|id_subject|

subjects_grades
|id_subject|id_grade|

students_grades
|id_student|id_grade|

任何帮助表示赞赏

我正在努力

SELECT students.name, subjects.name, grades.value
FROM students
INNER JOIN students_subjects
ON students.id_student = students_subjects.id_student
INNER JOIN subjects
ON subjects.id_subject = students_subjects.id_subject
INNER JOIN students_grades
ON students_grades.id_student = students.id_student
INNER JOIN grades
ON students_grades.id_grade = grades.id_grade
INNER JOIN subjects_grades
ON grades.id_grade = subjects_grades.id_grade

我得到下表

|     name |    name | value |
|----------|---------|-------|
|     Nico |  class1 |    70 |
|     Nico |  class1 |    40 |
|     Nico |  class2 |    70 |
|     Nico |  class2 |    40 |
|    Fadia |  class1 |    60 |
|    Fadia |  class1 |    55 |
| Cristian |  class2 |    50 |
| Cristian |  class2 |    40 |

但是如果我做AVG（grades.value）我只会得到第一行

最佳答案

只需添加一个组即可：

SELECT students.name, subjects.name, AVG(grades.value)
FROM students
    INNER JOIN students_subjects
    ON students.id_student = students_subjects.id_student
    INNER JOIN subjects
    ON subjects.id_subject = students_subjects.id_subject
    INNER JOIN students_grades
    ON students_grades.id_student = students.id_student
    INNER JOIN grades
    ON students_grades.id_grade = grades.id_grade
    INNER JOIN subjects_grades
    ON grades.id_grade = subjects_grades.id_grade
GROUP BY students.name, subjects.name

关于mysql - SQL选择学生的平均成绩，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/32018379/