Question

我在SQL Server 2005中有一个包含员工的层次结构表 - ＆gt;经理 - ＆gt;部门 - ＆gt;位置 - ＆gt;状态。

层次结构表的示例表：

ID   Name           ParentID   Type
1    PA             NULL       0 (group)
2    Pittsburgh     1          1 (subgroup)
3    Accounts       2          1 
4    Alex           3          2 (employee)
5    Robin          3          2
6    HR             2          1
7    Robert         6          2

第二个是事实表，其中包含员工薪资详细信息ID和薪水。

事实表的示例数据：

ID    Salary
4     6000
5     5000
7     4000

是否有任何方法可以显示层次结构表中的层次结构，并根据员工汇总工资总和。预期结果如

Name              Salary
PA                15000   (Pittsburgh + others(if any)) 
  Pittusburgh     15000   (Accounts + HR)
    Accounts      11000   (Alex + Robin)
      Alex         6000   (direct values)
      Robin        5000
    HR             4000
      Robert       4000

在我的生产环境中，层次结构表可能包含23000多行，而事实表可能包含300,000多行。因此，我考虑向查询提供任何级别的groupid以仅检索其子级及其相应的聚合值。有更好的解决方案吗？

Answer 1

此解决方案可生成正确的结果。只要索引到位，这应该对您的生产数据集执行正常，但我没有对您的示例数据进行测试。

DECLARE @tree TABLE
(ID INT
,name VARCHAR(15)
,ParentID INT
,TYPE TINYINT
)

DECLARE @salary TABLE
(ID INT
,Salary INT
)

INSERT @tree
      SELECT 1,'PA',NULL,0
UNION SELECT 2,'Pittsburgh',1,1
UNION SELECT 3,'Accounts',2,1
UNION SELECT 4,'Alex',3,2
UNION SELECT 5,'Robin',3,2
UNION SELECT 6,'HR',2,1
UNION SELECT 7,'Robert',6,2


INSERT @salary
      SELECT 4,6000
UNION SELECT 5,5000
UNION SELECT 7,4000


;WITH salaryCTE
AS
(
        SELECT t.*
               ,s.Salary
        FROM      @tree         AS t
        LEFT JOIN @salary       AS s
        ON        s.ID = t.ID
)
,recCTE
AS
(
        SELECT t.ID
               ,CAST(t.name AS VARCHAR(MAX)) AS name
               ,t.ParentID 
               ,ISNULL(t.Salary,0) AS Salary
               ,0 AS LEVEL
               ,CAST(t.ID AS VARCHAR(100)) AS ord
        FROM  salaryCTE   AS t
        WHERE t.ParentID IS NULL

        UNION ALL

        SELECT t.ID
               ,CAST(REPLICATE(' ',r.LEVEL) + t.name AS VARCHAR(MAX)) AS name
               ,t.ParentID
               ,ISNULL(t.Salary,0) AS Salary
               ,r.LEVEL + 1
               ,CAST(r.ord + '|' + CAST(t.ID AS VARCHAR(11)) AS VARCHAR(100)) AS ord
        FROM      salaryCTE     AS t        
        JOIN      recCTE        AS r
        ON        r.ID = t.ParentID
)
SELECT name
       ,salary
FROM (       
        SELECT r1.name
               ,r1.ord
               ,SUM(r2.salary) AS salary

        FROM recCTE AS r1
        LEFT JOIN recCTE AS r2
        ON   r2.ord LIKE r1.ord + '%'
        GROUP BY r1.name,r1.ord 
     ) AS x
ORDER BY ord,name

基于层次结构的聚合

1 个答案: