MYSQL：根据平均值对结果进行排序

Question

我有一张这样的表

------------------------------------------------------------
| id    | coursecodeid   | grade | workload  | easiness    |
------------------------------------------------------------
| 1     |  10            |  A+   |    5      |   4         |
| 2     |  10            |  A+   |    2      |   4         |
| 3     |  10            |  B    |    3      |   3         |
| 4     |  11            |  B+   |    2      |   3         |
| 5     |  11            |  A+   |    5      |   4         |
| 6     |  12            |  B    |    3      |   3         |
| 7     |  11            |  B+   |    7      |   8         |
| 8     |  11            |  A+   |    1      |   2         |
------------------------------------------------------------

我想根据每门课程的平均值计算结果（coursecodeid）。

平均值会是这样的

all workload values of the grade against courseid
+ all easiness values of the grade against courseid
/ 2 

从这个例子中coursecodeid = '10'有三个条目

avgworkload(5+ 2 + 3)/3=x 
avgeasiness(4 + 4 + 3)/3 = y
answer x+y/2 = z

所以我们必须像这样计算每门课程的平均值，并根据我们获得的平均值显示coursecodeids

和等级应根据特定课程的最大出现次数显示。

我正在运行此查询，但它没有按预期工作

SELECT 
 coursecode.id,
 feedback.grade CourseGrade,

 (( AVG( workload ) + AVG( easiness )) /2 ) as Avg
FROM coursecode
LEFT JOIN feedback ON feedback.coursecodeID = coursecode.id
GROUP BY coursecode.id
ORDER BY Avg DESC
LIMIT 11

Answer 1

试试这个

SELECT 
    (AVG(workload) + AVG(easiness)) / 2 AS result
FROM
    metagen.new_table
GROUP BY coursecodeid;