SQL 动态计算的百分比

Question

如何在SQL中动态计算百分比？

假设您有一个名为 Classes 的关注 table:

ClassSession       StudentName
---------------------------------
Evening            Ben
Morning            Chris
Afternoon          Roger
Evening            Ben
Afternoon          Ben
Morning            Roger
Morning            Ben
Afternoon          Chris

假设 Ben，我预计

Evening = 50 %
Afternoon = 25%
Morning = 25%

我期待克里斯

Morning = 50%
Afternoon = 50%
Evening = 0 %

所以ClassSession（三个session）应该是比较不变的

到目前为止，我已经尝试了以下 SQL 语句：

Select 
    ClassSession,
    (Count(ClassSession) * 100 / (Select Count(*) From Classes)) as Percentage
From 
    Classes
Where 
    StudentName = 'Chris'
Group By 
    ClassSession

Answer 1

一种方法使用条件聚合和 window 函数：

Select ClassSession,
       (sum(case when StudentName = 'Chris' then 100.0 else 0 end) /
        sum(sum(case when StudentName = 'Chris' then 100.0 else 0 end)) over ()
       ) as Percentage
From Classes
Group By ClassSession;

这将确保出现零的事件。

Answer 2

困难的部分是让零出现在给定会话中没有任何类的学生。

这是 PARTITION 外连接的作业。

select c.studentname, 
       s.classsession, 
       round(ratio_to_report(count(c.classsession)) 
            over ( partition by c.studentname),2) pct
from c partition by ( studentname ) 
   right outer join ( SELECT distinct classsession from c ) s 
          on s.classsession = c.classsession
group by c.studentname, s.classsession
order by c.studentname, s.classsession;

注意联接中的 PARTITION 关键字。这告诉 Oracle 为每个分区执行外部连接。因此，如果给定的 studentname 没有 classsession，请为该学生添加它。

此外，ratio_to_report 是计算百分比的好函数。

这是一个完整的示例，其中包含数据：

with c (ClassSession, StudentName) AS ( 
SELECT 'Evening',            'Ben' FROM DUAL UNION ALL
SELECT 'Morning',            'Chris' FROM DUAL UNION ALL
SELECT 'Afternoon',          'Roger' FROM DUAL UNION ALL
SELECT 'Evening',            'Ben' FROM DUAL UNION ALL
SELECT 'Afternoon',          'Ben' FROM DUAL UNION ALL
SELECT 'Morning',            'Roger' FROM DUAL UNION ALL
SELECT 'Morning',            'Ben' FROM DUAL UNION ALL
SELECT 'Afternoon',          'Chris' FROM DUAL)
select c.studentname, 
       s.classsession, 
       round(ratio_to_report(count(c.classsession)) 
            over ( partition by c.studentname),2) pct
from c partition by ( studentname ) 
   right outer join ( SELECT distinct classsession from c ) s on s.classsession = c.classsession
group by c.studentname, s.classsession
order by c.studentname, s.classsession;


╔══════════════════════════════════════════════════════════════════╗
║ STUDENTNAME CLASSSESSION PCT                                     ║
╠══════════════════════════════════════════════════════════════════╣
║ ----------- ------------ --------------------------------------  ║
║ Ben         Afternoon                                      0.25  ║
║ Ben         Evening                                         0.5  ║
║ Ben         Morning                                        0.25  ║
║ Chris       Afternoon                                       0.5  ║
║ Chris       Evening                                           0  ║
║ Chris       Morning                                         0.5  ║
║ Roger       Afternoon                                       0.5  ║
║ Roger       Evening                                           0  ║
║ Roger       Morning                                         0.5  ║
╚══════════════════════════════════════════════════════════════════╝

Answer 3

这是使用 SQL Server 2008 及更高版本的或多或少的传统方法。（在以后的版本中可能会有更方便的方法使用统计窗口函数来编写它。）

这将 return 至少一个 class 中所有学生的数据，所有 class 至少有一个学生的数据。如果表很大，取消注释 where 子句以一次获取一个学生的数据

一、测试数据：

CREATE TABLE #Test
 (
   ClassSession  varchar(20)  not null
  ,StudentName   varchar(20)  not null
 )

INSERT #Test values 
  ('Evening', 'Ben')
 ,('Morning', 'Chris')
 ,('Afternoon', 'Roger')
 ,('Evening', 'Ben')
 ,('Afternoon', 'Ben')
 ,('Morning', 'Roger')
 ,('Morning', 'Ben')
 ,('Afternoon', 'Chris')


SELECT *
 from #Test

以及查询：

WITH cteClasses
 as (--  First, get the list of classes
     SELECT distinct ClassSession
      from #Test
     )
 ,cteStudents
  as (--  Next, get a list of all students
      SELECT
         StudentName
        ,count(*) * 1.00  ClassCount
       from #Test
       --where StudenName = @StudentParameter
       group by StudentName
     )
 --  Mush them all together, and...
 SELECT
   st.StudentName
  ,cl.ClassSession
  ,count(te.StudentName) / st.ClassCount * 100  Percentage
 from cteStudents st
  cross join cteClasses cl
  left join #Test te
   on te.ClassSession = cl.ClassSession
    and te.StudentName = st.StudentName
 group by
   st.StudentName
  ,cl.ClassSession
  ,st.ClassCount
 order by 
   st.StudentName
  ,cl.ClassSession

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