如何获取mdx中相同成员的数量？

Question

我想确定两个日期之间的特定工作日（例如星期一，星期二等）的数量。 我认为类似以下内容的方法应该起作用，但是成员返回1。

我做错了什么？

WITH 
  MEMBER measures.NumberOfSameWeekDays AS 
    Count([Dim Date].[Day Of Week].CurrentMember) 
SELECT 
  measures.NumberOfSameWeekDays ON COLUMNS
 ,[Dim Date].[Day Of Week].[Day Of Week] ON ROWS
FROM [test]
WHERE 
  (
    [Dim Client].[Common Client UID].&[{ED8822E7-2873-4388-BC3A-CC553D939FC4}]
   ,
    [Dim Date].[Date Int].&[20150701] : [Dim Date].[Date Int].&[20150731]
  );

Answer 1

这是正在发生的事情的证明：

WITH 
  MEMBER measures.NumberOfSameWeekDays AS 
    Count([Date].[Day of Week].CurrentMember) 
  MEMBER measures.WeekDayCurrentMem AS 
    [Date].[Day of Week].CurrentMember.Member_Caption 
SELECT 
  {
    measures.NumberOfSameWeekDays
   ,measures.WeekDayCurrentMem
  } ON COLUMNS
 ,[Date].[Day of Week].[Day of Week] ON ROWS
FROM [Adventure Works]
WHERE 
    [Date].[Calendar].[Date].&[20050101]
  : 
    [Date].[Calendar].[Date].&[20050116];

这是以上结果：

这是上述行为的解决方案：

WITH 
  MEMBER measures.NumberOfSameWeekDays AS 
    Count
    (
      (EXISTING 
        [Date].[Day of Week].CurrentMember * [Date].[Calendar].[Date])
    ) 
SELECT 
  {
    measures.NumberOfSameWeekDays
  } ON COLUMNS
 ,[Date].[Day of Week].[Day of Week] ON ROWS
FROM [Adventure Works]
WHERE 
    [Date].[Calendar].[Date].&[20050101]
  : 
    [Date].[Calendar].[Date].&[20050131];

这将返回以下内容：

---

Sourav答案的简化版本-尽管仍然相当复杂-但由于它使用了迭代生成（Generate），因此可能会很慢：

WITH 
  MEMBER Measures.CountOfDays AS 
    Generate
    (
      (EXISTING 
        [Date].[Date].[Date].MEMBERS)
     ,[Date].[Day of Week]
     ,ALL
    ).Count 
SELECT 
  Measures.CountOfDays ON 0
 ,[Date].[Day of Week].[Day of Week].MEMBERS ON 1
FROM [Adventure Works]
WHERE 
  [Date].[Calendar].&[2005] : [Date].[Calendar].&[2006];

Answer 2

冒险作品版本：

WITH MEMBER Measures.CountOfDays  AS
GENERATE
    (
     EXISTING [Date].[Date].[Date].MEMBERS,
     EXISTING [Date].[Day of Week].[Day of Week].MEMBERS
     ,ALL
    ).COUNT


SELECT Measures.CountOfDays ON 0
,[Date].[Day of Week].[Day of Week].MEMBERS ON 1
FROM [Adventure Works]  
WHERE [Date].[Calendar].&[2005]: [Date].[Calendar].&[2006]

GENERATE部分会根据当前上下文并根据您可能拥有的任何过滤器来获取星期GENERATE 。