MDX 如何对集合进行分组

Question

有人想办法解决我的问题吗？

我必须比较不同时期的一些值（30.12.2017 – 05.1.2018 / 30.12.2016 – 05.01.2017 / ...） 因此，我定义了一个 mdx，每个周期都有集合：

WITH 
SET [Period 1]  as  { STRTOMEMBER( "[Created].[Year -  Month -  Date].[Date].&[2017-12-30T00:00:00]" 
                : STRTOMEMBER( "[Created].[Year -  Month -  Date].[Date].&[2018-01-04T00:00:00]" )  }

SET [Period 2]  as  { STRTOMEMBER( "[Created].[Year -  Month -  Date].[Date].&[2016-12-30T00:00:00]" ) 
                : STRTOMEMBER( "[Created].[Year -  Month -  Date].[Date].&[2017-01-04T00:00:00]" )  }


SELECT 
    NON EMPTY {  [Measures].[Average Price], [Measures].[Average Position]  } ON COLUMNS,

    NON EMPTY (  {   [Period 1], [Period 2] }  ) on rows

FROM [Cube]

当我想要每天的结果时，这很有效，但是如何获得每个时期的值（平均值/总和）？需要的结果是有 2 行，一个用于 Period1，一个用于 Period2。

Answer 1

您应该能够对 WITH 子句中的每个函数应用聚合函数：

WITH 
SET [Period 1]  as  
    STRTOMEMBER( "[Created].[Year -  Month -  Date].[Date].&[2017-12-30T00:00:00]" 
    : 
    STRTOMEMBER( "[Created].[Year -  Month -  Date].[Date].&[2018-01-04T00:00:00]" )
SET [Period 2]  as  
    STRTOMEMBER( "[Created].[Year -  Month -  Date].[Date].&[2016-12-30T00:00:00]" ) 
    : 
    STRTOMEMBER( "[Created].[Year -  Month -  Date].[Date].&[2017-01-04T00:00:00]" ) 
MEMBER [Created].[Year -  Month -  Date].[All].[Period 1 Agg] AS  
   AGGREGATE( [Period 1]    )   
MEMBER [Created].[Year -  Month -  Date].[All].[Period 2 Agg] AS  
   AGGREGATE( [Period 2]    )    
SELECT 
    NON EMPTY {  [Measures].[Average Price], [Measures].[Average Position]  } ON 0,
    NON EMPTY 
     {  
       [Created].[Year -  Month -  Date].[All].[Period 1 Agg],
       [Created].[Year -  Month -  Date].[All].[Period 2 Agg] 
     }  on 1
FROM [Cube];