2013-06-22 39 views
5

我有一個數據庫,它存儲組織的成本信息,按部門分解和按時間分階段進行。成本結構涉及親子關係;用戶可以在結構中的任何級別上指定成本值,唯一的限制是層次結構中較高級別的所有值都計算爲子節點的總和,如果任何子節點具有值;作爲子節點總和的父節點值不存儲在數據庫中。父子層次結構中的遞歸總和T-SQL

我需要一個查詢,將遞歸地計算基於他們的孩子的父母和沒有價值爲零將被設置(T-SQL,SQL 2008 R2)

[SQL小提琴] MS孩子的價值觀SQL Server 2008中架構設置

CREATE TABLE CostStructureNodes (
    Id INT NOT NULL PRIMARY KEY, 
    Name NVARCHAR(250) NOT NULL, 
    ParentNodeId INT, 
    FOREIGN KEY(ParentNodeId) REFERENCES CostStructureNodes(Id) 
); 

CREATE TABLE Years (
    Year INT NOT NULL PRIMARY KEY 
); 

CREATE TABLE CostsPerYear (
    NodeId INT NOT NULL, 
    Year INT NOT NULL, 
    Value DECIMAL(18,6) NOT NULL, 
    PRIMARY KEY(NodeId, Year), 
    FOREIGN KEY(NodeId) REFERENCES CostStructureNodes(Id), 
    FOREIGN KEY(Year) REFERENCES Years(Year) 
); 

INSERT INTO CostStructureNodes VALUES ('1', 'Total Costs', NULL); 
INSERT INTO CostStructureNodes VALUES ('2', 'R&D', 1); 
INSERT INTO CostStructureNodes VALUES ('3', 'Legal', 1); 
INSERT INTO CostStructureNodes VALUES ('4', 'HR', 1); 
INSERT INTO CostStructureNodes VALUES ('5', 'IT', 1); 
INSERT INTO CostStructureNodes VALUES ('6', 'Software', 5); 
INSERT INTO CostStructureNodes VALUES ('7', 'Hardware', 5); 

INSERT INTO Years VALUES (2010); 
INSERT INTO Years VALUES (2011); 
INSERT INTO Years VALUES (2012); 

INSERT INTO CostsPerYear VALUES (1, 2010, 100000); 
INSERT INTO CostsPerYear VALUES (2, 2011, 50000); 
INSERT INTO CostsPerYear VALUES (5, 2011, 20000); 
INSERT INTO CostsPerYear VALUES (6, 2012, 22000); 
INSERT INTO CostsPerYear VALUES (7, 2012, 13000); 
INSERT INTO CostsPerYear VALUES (2, 2012, 76000); 

鑑於以上的結構和樣本數據,這是怎麼會事的樣子:

|  NAME | YEAR | VALUE | 
    ------------------------------- 
    | Total Costs | 2010 | 100000 | 
    |   R&D | 2010 |  0 | 
    |   IT | 2010 |  0 | 
    | Software | 2010 |  0 | 
    | Hardware | 2010 |  0 | 
    |   HR | 2010 |  0 | 
    | Total Costs | 2011 | 70000 | 
    |   R&D | 2011 | 50000 | 
    |   IT | 2011 | 20000 | 
    | Software | 2011 |  0 | 
    | Hardware | 2011 |  0 | 
    |   HR | 2011 |  0 | 
    | Total Costs | 2012 | 111000 | 
    |   R&D | 2012 | 76000 | 
    |   IT | 2012 | 35000 | 
    | Software | 2012 | 22000 | 
    | Hardware | 2012 | 13000 | 
    |   HR | 2012 |  0 | 
+0

您可以重新檢查請求的輸出。鑑於HR的ID爲4,並且您在2012年的HR中插入了75000,在示例模式的最後一行中。我很難理解在要求的產出中,2012年人力資源的價值爲0。 – souplex

+0

我修復了示例數據插入。 – kjv

回答

4

這應該給一個正確的結果:

with DirectReport (ParentNodeId, Id, Name, Level, Struc, year) 
as 
(
    -- anchor 
    select a.ParentNodeId, a.Id, a.Name, 0 as Level, cast(':' + cast(a.Id as varchar) + ':' as varchar (100)) as Struc, y.year 
    from CostStructureNodes a, Years y 
    where a.ParentNodeId is null 
    union all 
    -- recursive 
    Select a.ParentNodeId, a.Id, a.Name, Level +1, cast(d.Struc + cast(a.Id as varchar)+ ':' as varchar(100)) as Struc, d.year 
    from CostStructureNodes a 
    join DirectReport d on d.Id = a.ParentNodeId 
) 

Select d.ParentNodeId, d.year, d.Id, d.Name, d.level, d.Struc,-- dd.Struc, 
sum(case when d.Struc = SUBSTRING(dd.Struc, 1, len(d.Struc))then c.Value else 0 end) as TotCost 
from DirectReport d 
    left join DirectReport dd on d.year = dd.year 
    join CostsPerYear c on c.Year = dd.year and c.NodeId = dd.Id 
group by d.ParentNodeId, d.year, d.Id, d.Name, d.level, d.Struc 
order by d.year, d.id 

這裏是小提琴鏈接:http://sqlfiddle.com/#!3/cd98d/22/0

注意兩個DirectReport部分之間的左連接,也爲保持部門沒有成本。

-1
WITH DirectReport (ParentNodeId, Id, Name, LEVEL, Struc) 
AS 
(
-- anchor 
SELECT a.ParentNodeId, a.Id, a.Name, 0 AS LEVEL, cast(':' + cast(a.Id AS varchar) + ':' AS varchar (100)) AS Struc 
FROM CostStructureNodes a 
WHERE a.ParentNodeId IS NULL 
UNION ALL 
-- recursive 
SELECT a.ParentNodeId, a.Id, a.Name, LEVEL +1, cast(d.Struc + cast(a.Id AS varchar)+ ':' AS varchar(100)) AS Struc 
FROM CostStructureNodes a 
    JOIN DirectReport d ON d.Id = a.ParentNodeId 
) 

SELECT d.ParentNodeId, d.Id, d.Name, d.level, d.Struc, 
sum(CASE WHEN d.Struc = SUBSTRING(dd.Struc, 1, len(d.Struc))THEN c.Value ELSE 0 END) AS TotCost 
FROM DirectReport d,DirectReport dd 
JOIN CostsPerYear c ON c.NodeId = dd.Id 
GROUP BY d.ParentNodeId,d.Id, d.Name, d.level, d.Struc 
ORDER BY d.id 
+0

http://sqlfiddle.com/#!3/bf0f3/16 – kamal