有没有办法可以缩短这个查询？答案

【问题标题】：Is there a way I can shorten this query?有没有办法可以缩短这个查询？
【发布时间】：2022-01-05 06:59:03
【问题描述】：

我正在解决一项任务，我设法找到了解决方案，但我不确定这是否是编写此查询的最佳方式

SELECT 
    students.studentname,
    CASE 
        WHEN (AVG(courses_student.exam_season_one) +  
              AVG(courses_student.exam_season_two) + 
              AVG(courses_student.degree_season_one) + 
              AVG(courses_student.degree_season_two)) / 4 >= 80 
            THEN 'EXCELLENT'
        WHEN (AVG(courses_student.exam_season_one) +  
              AVG(courses_student.exam_season_two) + 
              AVG(courses_student.degree_season_one) + 
              AVG(courses_student.degree_season_two)) / 4 >= 70 
            THEN 'VERY GOOD'
        WHEN (AVG(courses_student.exam_season_one) +  
              AVG(courses_student.exam_season_two) + 
              AVG(courses_student.degree_season_one) + 
              AVG(courses_student.degree_season_two)) / 4 >= 60 
            THEN 'GOOD'
        WHEN (AVG(courses_student.exam_season_one) +  
              AVG(courses_student.exam_season_two) + 
              AVG(courses_student.degree_season_one) + 
              AVG(courses_student.degree_season_two)) / 4 >= 50 
            THEN 'ACCEPTABLE'
        ELSE 'FAIL'
    END AS GRADE
FROM
    courses_student
JOIN 
    students ON students.student_id = courses_student.student_id
GROUP BY 
    students.studentname

如你所见，我重复了这个：

    WHEN (AVG(courses_student.exam_season_one) +  
          AVG(courses_student.exam_season_two) + 
          AVG(courses_student.degree_season_one) + 
          AVG(courses_student.degree_season_two)) / 4

四次！而且它看起来很乱，所以有没有一种方法可以让它更短，比如只写一次，然后只使用一个单词？（我试过用“AS”还是不行）

【问题讨论】：

在一行中有多个考试分数这一事实证明了数据设计存在缺陷。严格按照第三范式设计您的表格，您的问题将不复存在。

标签： sql oracle oracle-sqldeveloper

【解决方案1】：

为了清晰、测试和调试目的，您可以通过将相同的长表达式包含在子查询中来避免重复该表达式。例如：

select studentname,
  case when av >= 80 then 'EXCELLENT'
       when av >= 70 then 'VERY GOOD'
       when av >= 60 then 'GOOD'
       when av >= 50 then 'ACCEPTABLE'
       else 'FAIL'
  end as grade
from (
  SELECT s.studentname,
    (avg(cs.exam_season_one) + avg(cs.exam_season_two) + 
    avg(cs.degree_season_one) + avg(cs.degree_season_two)) / 4 as av
  FROM courses_student cs
  JOIN students c on s.student_id = cs.student_id
  GROUP BY s.studentname
) x

它看起来并不短，但发现错误并修复它要容易得多。

【讨论】：

【解决方案2】：

您可以像这样尝试 CTE：

With CTE( 
SELECT
    students.studentname, AVG(courses_student.exam_season_one) + AVG(courses_student.exam_season_two) + AVG(courses_student.degree_season_one) + AVG(courses_student.degree_season_two)) / 4 As average 
FROM
    courses_student 
    JOIN
        students 
        ON students.student_id = courses_student.student_id 
GROUP BY
    students.studentname
)
select
    studentname,
    case
        when
            average >= 80 
        then
            'EXCELLENT' 
        when
            average >= 70 
        then
            'VERY GOOD' 
        when
            average >= 60 
        then
            'GOOD' 
        when
            average >= 50 
        then
            'ACCEPTABLE' 
        else
            'FAIL' 
    end
    as grade 
from
    CTE

【讨论】：