0

我们有一个员工维度，其中包含我们在其上构建层次结构的经理（父子关系）的自我引用。

DimStaff 表：

 | SurrogateKey | BusinessKey | Employee Name | ManagerBusinessKey |  StartDate  |  EndDate  |
 |      1       |      1      |   Manager1    |        NULL        |  2013-01-01 | 2099-01-01|
 |      2       |      2      |   Manager2    |        NULL        |  2013-01-01 | 2099-01-01|
 |      3       |      3      |   Employee1   |        1           |  2013-01-01 | 2014-01-01|
 |      4       |      3      |   Employee1   |        2           |  2014-01-02 | 2099-01-01|

事实表：

 | StaffKey | DateKey  | Measure1 |
 |    3     | 20130405 | 10       |
 |    4     | 20140203 | 20       |

现在，以这个数据集为例，要求是

1-能够向下钻取层次结构

 Manager1
    ->   Employee1  
             ->   Measure1=10
 Manager2
    ->   Employee1  
             ->   Measure1=20

2- 选择一个人时聚合每个层次结构级别的值

Employee1    ->   Measure1=30

我们怎么能这样做呢？（问题是我们构建了它，但第二个要求不起作用，因为多维数据集接受 Employee1 的两个状态作为两个单独的实体并且不会聚合它们。）

4

2 回答 2

0

试试下面的 SQL 来得到答案

DECLARE @DimStaff table(
SurrogateKey INT,
BusinessKey INT,
EmployeeName Varchar(30),
ManagerBusinessKey INT, 
StartDate DATE,
EndDate DATE
);

Insert into @DimStaff
    (SurrogateKey, BusinessKey, EmployeeName, ManagerBusinessKey, StartDate, EndDate)
Values
    (1,1,'Manager1', NULL, '2013-01-01', '2099-01-01'),
    (2,2,'Manager2', NULL, '2013-01-01', '2099-01-01'),
    (3,3,'Employee1', 1, '2013-01-01', '2014-01-01'),
    (4,3,'Employee1', 2, '2014-01-02', '2099-01-01')

DECLARE @FactTable table(
StaffKey  INT,
DateKey   DATE,
Measure1  INT
);

INSERT INTO @FactTable 
    (StaffKey, DateKey, Measure1)
Values
    (3, '2013-04-05', 10 ),
    (4, '2013-02-03', 20 )

select t1.EmployeeName as Manager, t2.EmployeeName as Employee, ft.Measure1 as Measure from @DimStaff t1 
inner join @DimStaff t2 on t1.BusinessKey = t2.ManagerBusinessKey
inner join @FactTable ft on ft.StaffKey = t2.SurrogateKey

select t1.EmployeeName as Employee, sum (ft.Measure1) as TotalMeasure from @DimStaff t1 
inner join @FactTable ft on ft.StaffKey = t1.SurrogateKey
group by t1.EmployeeName
于 2014-07-21T23:50:16.290 回答
0

我听起来像您的员工姓名已在使用 SurrogateKey 作为 KeyColumns 属性的属性上定义。我将定义一个新属性，该属性具有 KeyColumns 的 BusinessKey 和 NameColumn 的 Employee Name。

于 2014-03-17T02:28:46.007 回答