假设我有三个成员的层次结构[Rank].[Rank]
[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。
但我认为ALLMEMBERS包含计算成员。谁能看到我做错了什么?
答案 0 :(得分:1)
你的命名有点混乱([Rank]。[Rank]。[Rank] vs [Rank]。[Rank]用法);你可以试试(注意名字中的额外[等级]):
with member [Rank].[Rank].[Rank].[Middle Managers] AS
[Rank].[Rank].&[Manager] + [Rank].[Rank].&[Supervisor]
[edit]附加你的计算结果。成员成为[Rank]工作级别以上的成员:
with member [Rank].[Rank].&[Boss].parent.[Middle Managers] AS
[Rank].[Rank].&[Manager] + [Rank].[Rank].&[Supervisor]