Сводная 2 столбца на основе 1 столбца

Question

У меня есть этот пример данных:

Я использую этот запрос:

SELECT        
    SDC.[Date],
    SDP.[Date] DateP,
    [ManagerProductName],
    SDC.[QuestionID], 
    SDC.QuestionValue [QuestionValueC],
    SDP.QuestionValue [QuestionValueP]
FROM
    [dbo].[SurveyData] SDC
INNER JOIN 
    [dbo].[SurveyData] SDP ON CAST(DATEADD(dd, -1, DATEADD(qq, DATEDIFF(qq, 0, SDC.Date), 0)) AS DATE) = SDP.date
                           AND SDC.QuestionID = SDP.QuestionID
                           AND SDC.SleeveID = SDP.SleeveID
INNER JOIN 
    [dbo].[FundManager] FM ON SDC.SleeveID = FM.SleeveID
INNER JOIN 
    [dbo].[Fund] F ON SDC.FundID = F.FundID

Но мне нужно преобразоватьэти данные примерно так:

Answer 1

Это был ответ:

SELECT   
               [Current Quarter]
              ,[Last Quarter]
              ,[FundName]
              ,[ManagerFullName]
              ,[ManagerProductName]
              ,[FileName], [1] [1 Current],  [11] [1 Previous],[2], [12]
from
(
  select 
               SDC.[Date] [Current Quarter]
              ,SDP.[Date] [Last Quarter]
              ,[FundName]
              ,[ManagerFullName]
              ,[ManagerProductName]
              ,SDC.[FileName], col, value
   FROM [dbo].[SurveyData] SDC
       INNER JOIN [dbo].[SurveyData] SDP
              ON CAST(DATEADD(dd, -1, DATEADD(qq, DATEDIFF(qq, 0, SDC.Date), 0)) AS DATE) = SDP.date
              AND SDC.QuestionID=SDP.QuestionID
              AND SDC.SleeveID=SDP.SleeveID
       INNER JOIN [dbo].[FundManager] FM
              ON SDC.SleeveID=FM.SleeveID
       INNER JOIN [dbo].[Fund] F
              ON SDC.FundID=F.FundID
  cross apply
  (
    select SDC.[QuestionValue], SDC.[QuestionID] union all -- AC 1/2
    select SDP.[QuestionValue], SDC.[QuestionID] + 10      -- BD 11/12
  ) c (value, col)
) d
pivot
(
  max(value)
  for col in ([1], [2], [11], [12])
) piv;