總和的平均值減最小值

Question

嘿傢伙，我有一個SQL語句，用於獲取不同活動類型（家庭作業，測驗等）的等級，如果該類型的下降最低，則會下降，否則它仍然存在。錯誤如下以及SQL代碼。

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 ... 

下面是我得到的錯誤：

- 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）。

事情是這樣的：

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 

設置與任何有意義的，你的情況劃分;我們無法分辨，因爲您在示例中省略了FROM ...。

然後，您可以在CASE語句中的任何計算中使用這些列別名。

Answer 2

如果你只需要去掉一個最低級（在領帶的情況下）

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 

如果您需要刪除所有最低等級：

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-運算符給出了一個結果（每個組）。最小運算符也是。那麼，avg-operator應該聚合什麼？