通过通用计算量度进行排序会导致无限递归吗? (MDX) [英] Ordering by a generic calculated measure results in infinite recursion? (MDX)

问题描述

我需要编写一个计算得出的度量，该度量应显示与任何给定维度的总和相比，Measure1的份额.我设法做到了:

I need to write a calculated measure that would show a share of Measure1 comparing to the total sum by any given dimension. I managed to do it like this:

CREATE MEMBER CURRENTCUBE.[Measures].[Test] AS
  [Measures].[Measure1] / ( AXIS( 1 ).ITEM( 0 ).DIMENSION.LEVELS( 0 ).ITEM( 0 ), [Measures].[Measure1] )

但是，由于[Test]的一般性质，因此无法通过此度量对尺寸进行排序.

But due to the generic nature of [Test], it is not possible to order a dimension by this measure.

SELECT [Measures].[Measure1] ON 0,
ORDER( [Dimension1].MEMBERS, [Measures].[Test] ) ON 1
FROM [MyCube]

执行上述代码会导致以下错误:

Executing the above code results in the following error:

检测到无限递归.依赖关系的循环是:测试->测试.

Infinite recursion detected. The loop of dependencies is: Test -> Test.

这很合逻辑-要获取AXIS( 1 ).ITEM( 0 )，我们需要在轴1上有一组维成员，但是只有在对这些成员按[Test]进行排序之后，才能获得该维成员.

Which is rather logical — to get AXIS( 1 ).ITEM( 0 ), we need a set of dimension members on axis 1, but this set is impossible to obtain until the members are sorted by [Test].

另一方面，如果我将[Test]定义为特定于某个尺寸，则它会按预期工作:

On the other hand, if I define [Test] as specific to some dimension, it works as expected:

CREATE MEMBER CURRENTCUBE.[Measures].[Test] AS
  [Measures].[Measure1] / ( [Dimension1].[All], [Measures].[Measure1] )

之所以可行，是因为[Dimension1].[All]寻址特定成员，而无需轴1首先评估其成员集.

It works, because [Dimension1].[All] addresses a particular member, without requiring axis 1 to evaluate its member set first.

是否有一种方法可以使此计算得出的度量通用?也许有可能以其他方式获取当前维度的[All]成员吗?

Is there a way to make this calculated measure generic? Maybe, it's possible to get current dimension's [All] member in another way?

推荐答案

如果仅显示一个度量，则可以重写第一个查询以避免错误:

If you are only displaying one measure, you could rewrite your first query to avoid the error:

SELECT [Measures].[Measure1] ON 0,
ORDER([Dimension1].MEMBERS) ON 1
FROM [MyCube]
WHERE ([Measures].[Test])

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