用于2个值列的SQL Pivot

Question

我需要透视这样的sql数据集。

studentID	year level	Results	Dimension
23096	7	576	reading
23096	7	640	writing
23096	7	704	spelling
23096	7	768	numeracy
23096	7	832	grammarPunctuation
23096	8	896	reading
23096	8	960	writing
23096	8	1024	spelling
23096	8	1088	numeracy
23096	8	1152	grammarPunctuation
23104	8	1216	reading
23104	8	1280	writing
23104	8	1344	spelling
23104	8	1408	numeracy
23104	8	1472	grammarPunctuation
23116	8	1536	reading
23116	8	1600	writing
23116	8	1664	spelling
23116	8	1728	grammarPunctuation
23117	7	1792	reading
23117	7	1856	writing
23117	7	1920	spelling
23117	7	1984	numeracy
23117	7	2048	grammarPunctuation

看起来像

StudentID Dimension   Year7_Results  Year8_Results  
 23096    reading             576                896     
 23096    writing             640                960 
 23096    spelling            704                1024
 23096    numeracy            768                1088
 23096    grammarPunctuation  832                1152

我的代码

Select * from 
(
select StudentID,yearlevel,resultes,Dimantion  ,ROW_NUMBER() OVER(partition by studentID,Dimension order by studentID,Dimension) AS PX,
'R' + CAST(ROW_NUMBER() OVER(partition by studentID,Dimension order by studentID,Dimension) as varchar(2) ) AS PX2
 from table
 )Temp
 PIVOT
 (
 MAX(Dimension) FOR PX IN ("1","2")
 )PVI1 
 PIVOT 
 (
 SUM(resultes) FOR PX2 IN ("R1","R2")
 )PIVOT2

Answer 1

您可以使用条件聚合来代替数据透视：

数据示例：

declare @table as table (
       studentID  int          not null,
       yearlevel  int          not null,
       results    int          null,
       dimension  nvarchar(50) null
    )
    
    insert into @table
    values 
    (23096,7,576,'reading'),
    (23096,7,640,'writing'),
    (23096,7,704,'spelling'),
    (23096,7,768,'numeracy'),
    (23096,7,832,'grammarPunctuation'),
    (23096,8,896,'reading'),
    (23096,8,960,'writing'),
    (23096,8,1024,'spelling'),
    (23096,8,1088,'numeracy'),
    (23096,8,1152,'grammarPunctuation'),
    (23104,8,1216,'reading'),
    (23104,8,1280,'writing'),
    (23104,8,1344,'spelling'),
    (23104,8,1408,'numeracy'),
    (23104,8,1472,'grammarPunctuation'),
    (23116,8,1536,'reading'),
    (23116,8,1600,'writing'),
    (23116,8,1664,'spelling'),
    (23116,8,1728,'grammarPunctuation'),
    (23117,7,1792,'reading'),
    (23117,7,1856,'writing'),
    (23117,7,1920,'spelling'),
    (23117,7,1984,'numeracy'),
    (23117,7,2048,'grammarPunctuation')

查询：

select  StudentID, 
        Dimension, 
        max(case when yearlevel = 7 then results end) as Year7_Results  ,
        max(case when yearlevel = 8 then results end) as Year8_Results  
from  @table
group by StudentID, Dimension

Answer 2

您可以在不使用数据透视功能的情况下解决该问题。首先创建包含所有学生记录的表。

drop table if exists #student_results

create table #student_results (
   studentID  int          not null,
   year_level int          not null,
   Results    int          null,
   Dimension  nvarchar(50) null
)

然后从示例数据中插入一些结果。

insert into #student_results
values 
(23096, 7,  576, 'reading' )
, (23096,   7,  640, 'writing' )
, (23096,   7,  704, 'spelling' )
, (23096,   7,  768, 'numeracy' )
, (23096,   8,  960, 'writing')
, (23096,   8,  1024, 'spelling')
, (23096,   8,  1088, 'numeracy')
, (23096,   8,  1152, 'grammarPunctuation')
, (23104,   8,  1216, 'reading')
, (23104,   8,  1280, 'writing')
, (23104,   8,  1344, 'spelling')
, (23104,   8,  1408, 'numeracy')
, (23104,   8,  1472, 'grammarPunctuation')
, (23116,   8,  1536, 'reading')
, (23116,   8,  1600, 'writing')
, (23116,   8,  1664, 'spelling')
, (23116,   8,  1728, 'grammarPunctuation')
, (23117,   7,  1792, 'reading')
, (23117,   7,  1856, 'writing')
, (23117,   7,  1920, 'spelling')
, (23117,   7,  1984, 'numeracy')
, (23117,   7,  2048, 'grammarPunctuation')
;

使用CTE，我们打破了逻辑。

1. 我们创建一个包含StudentID和Dimension的所有组合的表格。
2. 使用7年级的结果创建表格
3. 使用8年级的结果创建表格
4. 合并结果

with students
    as (select studentID
            , Dimension
        from   #student_results
        group by studentID
             , Dimension),
    year7
    as (select *
        from   #student_results
        where  year_level = 7),
    year8
    as (select *
        from   #student_results
        where  year_level = 8)
    select students.studentID
        , students.Dimension
        , year7.Results as Year7_Results
        , year8.Results as Year8_Results
    from   students
    left join year7
            on students.studentID = year7.studentID
              and students.Dimension = year7.Dimension
    left join year8
            on students.studentID = year8.studentID
              and students.Dimension = year8.Dimension;