Question

+------------+-------+---------+---------------+
| STudent_ID | Marks | Subject | EntryPoints   |
+------------+-------+---------+---------------+
|          1 |    50 | Maths   |          10   |
|          2 |    50 | Maths   |          10   |
|          3 |    45 | Maths   |          10   |
|          1 |    30 | History |          20   |
|          2 |    30 | History |          20   |
|          3 |    30 | History |          20   |
+------------+-------+---------+---------------+

预期产出：

+------------+-------+---------+---------------+ 
| student_id | Marks | Subject | TotalPoints   | 
+------------+-------+---------+---------------+ 
|          1 |    50 | Maths   | 5             | 
|          2 |    50 | Maths   | 5             | 
|          1 |    30 | History | 6.66          | 
|          2 |    30 | History | 6.66          | 
|          3 |    30 | History | 6.66          | 
+------------+-------+---------+---------------+

总积分计算

对于数学，入学点数为10，学生得分最大数为2，因此10/2 = 5 历史记录入口点数为20，学生得分最大值为3，因此20/3 = 6.66

我试过查询：

select student_id,marks,subject from
(
select student_id,marks,subject,dense_rank() over ( partition by subject order by marks desc) rn  from test
) t
where rn=1

输出：

+------------+-------+---------+ 
| Student_id | Marks | Subject | 
+------------+-------+---------+ 
|          1 |    50 | Maths   | 
|          2 |    50 | Maths   | 
|          1 |    30 | History | 
|          2 |    30 | History | 
|          3 |    30 | History | 
+------------+-------+---------+

我无法获得如何获取查询中的总分数列

Answer 1

我假设最高分是所有具有相同主题的分数中的最大分数。

SELECT 
    Student_ID
    ,Marks
    ,T.Subject
    ,CONVERT(decimal(18, 2), 
       CONVERT(float, EntryPoints)/
       CONVERT(float, COUNT(*) OVER(PARTITION BY T.Subject))) as 'TotalPoints'
  FROM @T T
  INNER JOIN (SELECT DISTINCT Subject, MAX(Marks) OVER(PARTITION BY Subject) 
  as Max_Marks FROM @T) Scores
  ON T.Subject = Scores.Subject

  WHERE Marks = Scores.Max_Marks
  ORDER BY Marks DESC, Student_ID

Answer 2

您可以在查询中使用Group by获取主题计数，然后使用相同的查询join结果。

如下： -

Select a.student_id,a.marks,a.subject, 
Cast(a.EntryPoints as decimal ) / cast(b.CountPerSubject as decimal) TotalPoints   
from 
(
select student_id,marks,subject ,EntryPoints
    from
    (
            select student_id,marks,
            subject, EntryPoints,
            dense_rank() over ( partition by subject order by marks desc) rn  

            from test
    ) t
where rn=1
) a

Left Join
(
    select subject, count(subject) CountPerSubject
    from
    (
    select student_id,marks,subject ,EntryPoints
    from
    (
            select student_id,marks,
            subject, EntryPoints,
            dense_rank() over ( partition by subject order by marks desc) rn  

            from test
    ) t
where rn=1 ) c

group by subject) b
on a.subject =  b.subject

需要首先分配用户数量的入口点

2 个答案: