假设我有[Rank].[Rank]
三个成员的层次结构
[Rank].[Rank].&[Boss]
[Rank].[Rank].&[Manager]
[Rank].[Rank].&[Supervisor]
[Rank].[Rank].&[Serf]
然后在查询中创建一个计算成员
MEMBER [Rank].[Rank].[Middle Managers] AS
[Rank].[Rank].&[Manager] + [Rank].[Rank].&[Supervisor]
如果我说
with
member [Rank].[Rank].[Middle Managers] AS
[Rank].[Rank].&[Manager] + [Rank].[Rank].&[Supervisor]
select
{
[Measures].[Hours]
}
on 0
, [Rank].[Rank].[Rank].ALLMEMBERS
on 1
from
some_cube
我没有得到 [Rank].[Rank].[Middle Managers] 出现在结果集中,但如果我使用
with
member [Rank].[Rank].[Middle Managers] AS
[Rank].[Rank].&[Manager] + [Rank].[Rank].&[Supervisor]
select
{
[Measures].[Hours On Stack Overflow]
}
on 0
, [Rank].[Rank].[Rank].ALLMEMBERS + [Rank].[Rank].[Middle Managers]
on 1
from
some_cube
我明白了。
但我的印象是 ALLMEMBERS 包括计算成员。谁能看到我做错了什么?