强制计算成员在聚集之前在叶级别进行评估

Question

我找到了此解决方案：

    with calculated member [Measures].[Test1] 
as 
 iif (isleaf([Time].[Calendar].currentmember)), 
[Measures].[Measure1]/[Measures].[Measure2], 
Sum([Time].[Calendar].currentmember.children, [Measures].[Test1]))

    select Test1 on 0,
    [Time].[Calendar]. members on 1
    from 
    [MyCube]

它有效。

但是我想知道是否可以具有两个或更多个维度，并编写如下内容：

    with calculated member [Measures].[Test1] as 
    iif (isleaf([Time].[Calendar].currentmember)) and  isleaf([SecondDimension].[SecondDimensionHierarchy].currentmember), 
[Measures].[Measure1]/[Measures].[Measure2], 
**Sum([Time].[Calendar].currentmember.children, [Measures].[Test1])**)

   select Test1 on 0,
    [Time].[Calendar]. members on 1
    from 
    [MyCube]

我尝试了此操作，但叶子得到了正确的结果，但是高级成员的值为空-我有两个层次结构时，我不知道如何写这个总和Sum([Time].[Calendar].currentmember.children, [Measures].[Test1])。可能吗？

Answer 1

我想我已经找到了解决方法：

with calculated member [Measures].[Test1] as 
    iif (isleaf([Time].[Calendar].currentmember)) and  isleaf([SecondDimension].[SecondDimensionHierarchy].currentmember), 
[Measures].[Measure1]/[Measures].[Measure2], 
Sum((DESCENDANTS([Time].[Calendar].currentmember, , LEAVES) * DESCENDANTS([SecondDimension].[SecondDimensionHierarchy].currentmember, , LEAVES), [Measures].[Test1]))

   select Test1 on 0,
    [Time].[Calendar]. members on 1
    from 
    [MyCube]