SQL 通过调查了解课程受欢迎程度

Question

我正在尝试编写一个显示课程受欢迎程度的 SQL 查询，按降序排列。

课程受欢迎程度以分数衡量，其确定如下：对于每个调查：
一个。如果票数差值 > 总票数的10%，受欢迎的课程得1分，不受欢迎的课程得0分
b。如果票数差异 <= 总票数的 10%，则每个课程获得 0.5 点

course_id	course_name	faculty
1001	economics_101	business
1002	algebra_101	math
1003	geometry_101	math
1004	management_101	business
1005	marketing_101	business
1006	physics_101	science

survey_id	option_a	option_b	votes_a	votes_b
2001	economics_101	geometry_101	61	34
2002	algebra_101	economics_101	31	68
2003	marketing_101	management_101	11	72
2005	management_101	algebra_101	43	54
2004	geometry_101	marketing_101	48	46

目前取得的成绩：

course	popularity
economics_101	4
management_101	2
algebra_101	2
marketing_101	1
geometry_101	1
[NULL]	0

到目前为止我设法加入了它，如果能提供有关优化此查询的意见，那就太好了：

    WITH x AS 
(
WITH b AS 

(

WITH a as 
(
select * from course c
LEFT  JOIN survey s
on c.course_name = s.option_a 

UNION ALL 

select * from course c
LEFT  JOIN survey s
on c.course_name = s.option_b
)

SELECT a.*,
SUM(votes_a+votes_b) as total_votes,
CASE WHEN (a.votes_a - a.votes_b) > (0.1*SUM(votes_a+votes_b))  THEN 1 
     WHEN (a.votes_b - a.votes_a) <= (0.1*SUM(votes_a+votes_b))  THEN 0.5
     ELSE 0
END AS 'Popularity_a',

CASE WHEN (a.votes_b - a.votes_a) > (0.1*SUM(votes_a+votes_b))  THEN 1 
     WHEN (a.votes_a - a.votes_b) <= (0.1*SUM(votes_a+votes_b))  THEN 0.5
     ELSE 0
END AS 'Popularity_b'


FROM 
a
GROUP BY 
a.course_name ,
a.course_id,
a.faculty ,
a.survey_id ,
a.option_a ,
a.option_b ,
a.votes_a ,
a.votes_b 
)

SELECT b.option_a  as course,
b.Popularity_a as pop
FROM b
LEFT JOIN 
course cx 
ON b.option_a  = cx.course_name  

UNION ALL 

SELECT b.option_b as course ,
b.Popularity_b as pop
FROM b
LEFT JOIN 
course cx 
ON b.option_b  = cx.course_name 
)

select
x.course ,
sum (x.pop) as popularity
from x
GROUP BY
x.course 

order by popularity desc

Answer 1

使用UNION ALL提取所有课程以及他们从tablesurvey中获得的相应分数并聚合以获得popularity。
然后加入 course:

WITH 
  cte AS (
    SELECT option_a course_name, 
           CASE 
             WHEN votes_a - votes_b > 0.1 * (votes_a + votes_b) THEN 1.0 
             WHEN votes_b - votes_a > 0.1 * (votes_a + votes_b) THEN 0.0
             ELSE 0.5 
           END points
    FROM survey
    UNION ALL
    SELECT option_b, 
           CASE 
             WHEN votes_b - votes_a > 0.1 * (votes_a + votes_b) THEN 1.0 
             WHEN votes_a - votes_b > 0.1 * (votes_a + votes_b) THEN 0.0
             ELSE 0.5 
           END
    FROM survey
  ),
  points AS (
    SELECT course_name, SUM(points) total_points
    FROM cte
    GROUP BY course_name
  )
SELECT c.*, COALESCE(p.total_points, 0) popularity
FROM course c LEFT JOIN points p
ON p.course_name = c.course_name
ORDER BY popularity DESC;

参见demo。

SQL 通过调查了解课程受欢迎程度

SQL to find course popularity from a survey

sql

sqlite

group-by

case

sql-order-by