MDX中作为度量值的维度成员
我需要返回一个维度成员作为度量值 鉴于: 尺寸MDX中作为度量值的维度成员,mdx,Mdx,我需要返回一个维度成员作为度量值 鉴于: 尺寸 [Customer].[customer name] members {ACME, EMCA, EMC} [Salesperson].[person name] members {Bob, Fred, Mary, Joe} with the property "name" 措施 价值观 关系 The Customer is a dimension of the facts that contain Value The Salespers
[Customer].[customer name] members {ACME, EMCA, EMC}
[Salesperson].[person name] members {Bob, Fred, Mary, Joe} with the property "name"
措施
价值观
关系
The Customer is a dimension of the facts that contain Value
The Salesperson is a dimensions of the facts that contain Value, each customer has one salesperson associated
我正在努力做到以下几点:
创建将在列中为客户返回销售人员姓名的度量值。e、 g
| Customer |Sales person | Value |
| ACME | Bob | 500 |
| EMCA | Bob | 540 |
| EMC | Mary | 840 |
我试过这样做:
成员[measure].[sp_name]为[saleperson].[person name].currentmember.properties(“name”)
挑选
0上的{[measure].[sp_name],[measures].[value]}
{[customer].[customer name].members}在1上
从
但它总是返回错误“属性名称对[Salesperson].[all Salesperson]无效”
如果我将与成员[measure].[sp_name]一起用作[saleperson].[person name].currentmember.firstindex.properties(“name”)
我为所有的顾客安排了一个人,这显然是不对的。我遗漏了什么吗?我想说的是,在Analysis Services 2008中,这应该在您实现它时起作用,为
所有成员返回null
。如果您使用的环境更敏感,那么您可以使用
With Member [measure].[sp_name] as
IIF([Salesperson].[person name].currentmember IS [Salesperson].[all salesperson],
'All',
[Salesperson].[person name].properties("name")
)
Select {[measure].[sp_name], [measures].[value]} on 0,
{[customer].[customer name].members} on 1
from [MyCube]
[Salesperson].[person name].CurrentMember.Member_Name