查找頂級父ID

Question

在一個表中我有兩列像下面

ID ParentID 
1 0x0 
2 1 
3 2 
9 0x0 
5 9 
6 5 
25 0x0 
30 25 

如何獲得頂級父ID
結果應該像

ID TopParentID 
3 1 
6 9 
30 25 

Answer 1

另一個短CTE：

WITH cte AS(
     SELECT *, id AS topparent 
     FROM t 
     WHERE parentid IS NULL 
    UNION ALL 
     SELECT t.*, c.topparent 
     FROM t JOIN cte c ON c.id = t.parentid 
     WHERE t.id <> t.parentid 
) 
SELECT * FROM cte 

這裏測試：SQL Fiddle

---

編輯：這該死的SQLfiddle沒有響應一次。下面

CREATE table t (id INT, parentid INT); 
INSERT INTO T VALUES (1 , NULL); 
INSERT INTO T VALUES (2 , 1 );   
INSERT INTO T VALUES (3 , 2 );   
INSERT INTO T VALUES (9 , NULL);   
INSERT INTO T VALUES (5 , 9 );   
INSERT INTO T VALUES (6 , 5 );   
INSERT INTO T VALUES (25 , NULL);   
INSERT INTO T VALUES (30 , 25 ); 

Answer 2

您可以使用遞歸CTE：

WITH CTE AS 
(
    SELECT *, 1 [Level] 
    FROM dbo.YourTable 
    UNION ALL 
    SELECT A.ID, B.[Parent ID], [Level] + 1 
    FROM CTE A 
    INNER JOIN dbo.YourTable B 
     ON A.[Parent ID] = B.Id 
), CTE2 AS 
(
    SELECT *, RN = ROW_NUMBER() OVER(PARTITION BY ID ORDER BY [Level] DESC) 
    FROM CTE 
    WHERE [Parent ID] <> 0 
) 
SELECT ID, [Parent ID] [Top Level Parent Id] 
FROM CTE2 
WHERE RN = 1 
OPTION (MAXRECURSION 0) 
; 

Here is a link with a demo。

的結果是：

╔════╦═════════════════════╗ 
║ ID ║ Top Level Parent Id ║ 
╠════╬═════════════════════╣ 
║ 2 ║     1 ║ 
║ 3 ║     1 ║ 
║ 5 ║     9 ║ 
║ 6 ║     9 ║ 
║ 30 ║     25 ║ 
╚════╩═════════════════════╝ 

Answer 3

測試DDL代碼如果頂級節點總是跟隨到單個葉節點的單個路徑，則所期望的結果，可以通過組合麪包屑和遞歸的CTE實現。以下示例演示了該方法：

CREATE TABLE #Items 
(
    ID    INT 
    ,[ParentID]  INT 
) 

INSERT INTO #Items 
VALUES 
(1,  0x0), 
(2,  1), 
(3,  2), 
(9,  0x0), 
(5,  9), 
(6,  5), 
(25, 0x0), 
(30, 25) 



;WITH CTE_Breadcrumbs 
AS 
(
    SELECT  ID 
       ,ParentID 
       ,CAST(ID AS NVARCHAR(1000)) AS [Path] 
       ,LEN(CAST(ID AS NVARCHAR(1000))) AS [PathLength] 
       ,ID AS [RootNodeID] 
    FROM  #Items 
    WHERE  ParentID = 0x0 
    UNION ALL 
    SELECT  Child.ID 
       ,Child.ParentID 
       ,CAST(Parent.[Path] + N'\' + CAST(Child.ID AS NVARCHAR(1000)) AS NVARCHAR(1000)) AS [Path] 
       ,LEN(CAST(Parent.[Path] + N'\' + CAST(Child.ID AS NVARCHAR(1000)) AS NVARCHAR(1000))) AS [PathLength] 
       ,CAST(
        CASE 
         WHEN CHARINDEX('\', [Path]) > 0 THEN LEFT([Path], CHARINDEX('\', [Path]) - 1) 
         ELSE Child.ParentID 
        END 
       AS INT) AS [RootNodeID] 
    FROM  #Items Child 
    INNER JOIN CTE_Breadcrumbs Parent ON Child.ParentID = Parent.ID 
) 
, 
CTE_FullPathID 
AS 
(
    SELECT  [RootNodeID] 
       ,MAX([PathLength]) AS [PathFullLength] 
    FROM  CTE_Breadcrumbs 
    GROUP BY [RootNodeID] 

) 
SELECT  CTE_Breadcrumbs.RootNodeID 
      ,CAST(
       REVERSE(
          CASE 
           WHEN CHARINDEX('\', REVERSE([Path])) > 0 THEN LEFT(REVERSE([Path]), CHARINDEX('\', REVERSE([Path])) - 1) 
           ELSE NULL 
          END 
         ) 
      AS INT) AS [LeafNodeID] 
      ,CTE_Breadcrumbs.[Path] 
FROM  CTE_Breadcrumbs 
INNER JOIN CTE_FullPathID ON CTE_Breadcrumbs.[RootNodeID] = CTE_FullPathID.[RootNodeID] AND CTE_Breadcrumbs.PathLength = CTE_FullPathID.[PathFullLength]