总和的平均值减最小值

Question

嘿家伙，我有一个SQL语句，用于获取不同活动类型（家庭作业，测验等）的等级，如果该类型的下降最低，则会下降，否则它仍然存在。错误如下以及SQL代码。

SELECT  Student.firstName, Student.lastName, 'Grades' = 
      CASE 
       WHEN Grades.activityType = 'Homework' THEN 
       CASE WHEN Policy.drop_hw = 1 THEN 
        (AVG(SUM(Grades.grade) - MIN(Grades.grade))) * (Policy.homework/100) 
       ELSE 
        (AVG(Grades.grade) * (Policy.homework/100)) 
       END 
      END, Course.courseNum, Course.sectNum, Grades.activityType 

FROM ... 

下面是我得到的错误：

- Cannot perform an aggregate function on an expression containing an aggregate or a subquery. 
- Column 'Policy.drop_hw' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause. 

Answer 1

研究分析功能。 （SO question,Oracle documentation）。

事情是这样的：

AVG(Grades.grade) OVER (PARTITION BY Grades.student_id) AS avg_of_grades 

和：

(AVG(SUM(Grades.grade) - MIN(Grades.grade))) OVER (PARTITION BY Grades.student_id) AS avg_grades_with_drop 

设置与任何有意义的，你的情况划分;我们无法分辨，因为您在示例中省略了FROM ...。

然后，您可以在CASE语句中的任何计算中使用这些列别名。

Answer 2

如果你只需要去掉一个最低级（在领带的情况下）

SELECT student_id, AVG(grade) 
FROM (
     SELECT *, ROW_NUMBER() OVER (PARTITION BY student_id ORDER BY grade) rn 
     FROM my_tables 
     ) 
WHERE NOT (drop_hw = 1 AND rn = 1) 
GROUP BY 
     student_id 

如果您需要删除所有最低等级：

SELECT student_id, AVG(grade) 
FROM (
     SELECT *, MIN(grade) OVER (PARTITION BY student_id) mingrade 
     FROM my_tables 
     ) 
WHERE NOT (drop_hw = 1 AND grade = mingrade) 
GROUP BY 
     student_id 

Answer 3

su m-运算符给出了一个结果（每个组）。最小运算符也是。那么，avg-operator应该聚合什么？