SUM和多个IIF功能状况MDX

Question

我已经写了下面的MDX查询这里我在做什么尝试获得基于在IIF功能应用多个条件汤姆的结果：

WITH 
    SET [kpi_study] AS 
    {[study].[study].[BHC June12]} 
    SET [geographic] AS 
    {[territory.market_hierarchy].[state].[MP]} 
    SET [brand] AS 
    {[brand.brand_hierarchy].[brand].[Gold Flake (Unspecified)]} 
    SET [edu12] AS 
    IIF 
    (
     'All' = 'All' 
    ,[education].[education].MEMBERS 
    ,[education].[education].[All] 
    ) 
    SET [town] as 
    IIF(
     'All' = 'All' 
     ,[territory.market_hierarchy].[town_class].MEMBERS 
     ,[territory.market_hierarchy].[town_class].[All] 
     ) 
    SET [occp] as 
    IIF(
     'All' = 'All' 
      ,[occupation].[occupation].MEMBERS 
      ,[occupation].[occupation].[All] 
     ) 
    MEMBER [Measures].[t] AS 
    SUM(([edu12],[town],[occp]),[Measures].[tom]) 
SELECT 
    NON EMPTY 
    {[Measures].[t]} ON COLUMNS 
FROM [funnel_analysis] 
WHERE 
    {[kpi_study]*[geographic]*[brand]} 

但收到的一些错误。对于单一的iif功能，它的工作正常，即：**(SUM([edu12],[Measures].[tom]))**无法找出我在哪里做错了多个。

Answer 1

我会做一个明确的交叉连接。 也请摆脱你正在创建的那些单一成员自定义集 - 这不是标准做法 - 只要把它们直接放在你的WHERE条款中即可。

WITH 
    SET [edu12] AS 
    IIF(
    'All' = 'All' 
    ,{[education].[education].MEMBERS} 
    ,[education].[education].[All] 
    ) 
    SET [town] as 
    IIF(
     'All' = 'All' 
     ,{[territory.market_hierarchy].[town_class].MEMBERS} 
     ,[territory.market_hierarchy].[town_class].[All] 
    ) 
    SET [occp] as 
    IIF(
     'All' = 'All' 
     ,{[occupation].[occupation].MEMBERS} 
     ,[occupation].[occupation].[All] 
    ) 
    MEMBER [Measures].[t] AS 
    SUM(
     [edu12] 
     *[town] 
     *[occp] 
     ,[Measures].[tom] 
    ) 
SELECT 
    NON EMPTY 
    {[Measures].[t]} ON COLUMNS 
FROM [funnel_analysis] 
WHERE 
    (
    [study].[study].[BHC June12] 
    ,[territory.market_hierarchy].[state].[MP] 
    ,[brand.brand_hierarchy].[brand].[Gold Flake (Unspecified)] 
) 

我更愿意尝试类似的使用Aggregate如下：

WITH 
    MEMBER [education].[education].[All].[edu12] AS 
    AGGREGATE(IIF(
    'All' = 'All' 
    ,{[education].[education].MEMBERS} 
    ,[education].[education].[All] 
    )) 
    MEMBER [territory.market_hierarchy].[town_class].[All].[town] as 
    AGGREGATE(IIF(
     'All' = 'All' 
     ,{[territory.market_hierarchy].[town_class].MEMBERS} 
     ,[territory.market_hierarchy].[town_class].[All] 
    )) 
    MEMBER [occupation].[occupation].[All].[occp] as 
    AGGREGATE(IIF(
     'All' = 'All' 
     ,{[occupation].[occupation].MEMBERS} 
     ,[occupation].[occupation].[All] 
    )) 
    MEMBER [Measures].[t] AS 
    (
     [education].[education].[All].[edu12] 
     ,[territory.market_hierarchy].[town_class].[All].[town] 
     ,[occupation].[occupation].[All].[occp] 
     ,[Measures].[tom] 
    ) 
SELECT 
    NON EMPTY 
    {[Measures].[t]} ON COLUMNS 
FROM [funnel_analysis] 
WHERE 
    (
    [study].[study].[BHC June12] 
    ,[territory.market_hierarchy].[state].[MP] 
    ,[brand.brand_hierarchy].[brand].[Gold Flake (Unspecified)] 
) 

探索脚本示例 - 这是否给你你所期望？如果没关系，请继续阅读更复杂脚本的其他部分：

WITH 
    SET [edu12] AS 
    IIF(
    'All' = 'All' 
    ,{[education].[education].MEMBERS} 
    ,[education].[education].[All] 
    ) 
SELECT 
    [edu12] ON ROWS, 
{[Measures].[tom]} ON COLUMNS 
FROM [funnel_analysis] 
WHERE 
    (
    [study].[study].[BHC June12] 
    ,[territory.market_hierarchy].[state].[MP] 
    ,[brand.brand_hierarchy].[brand].[Gold Flake (Unspecified)] 
)