SQL SERVER 2008 CTE生成结点的FullPath

数据库  /  管理员 发布于 8年前   244

好的，现在来看如何生成FullPath:
复制代码 代码如下:
DECLARE @tbl TABLE
(
Id int
,ParentId int
)
INSERT INTO @tbl
( Id, ParentId )
VALUES ( 0, NULL )
, ( 8, 0 )
, ( 12, 8 )
, ( 16, 12 )
, ( 17, 16 )
, ( 18, 17 )
, ( 19, 17 )

WITH abcd
AS (
-- anchor
SELECT id
,ParentID
,CAST(id AS VARCHAR(100)) AS [Path]
FROM @tbl
WHERE ParentId IS NULL
UNION ALL
--recursive member
SELECT t.id
,t.ParentID
,CAST(a.[Path] + ',' + CAST( t.ID AS VARCHAR(100)) AS varchar(100)) AS [Path]
FROM @tbl AS t
JOIN abcd AS a ON t.ParentId = a.id
)
SELECT Id ,ParentID ,[Path]
FROM abcd
WHERE Id NOT IN ( SELECT ParentId
FROM @tbl
WHERE ParentId IS NOT NULL )

返回：
Id ParentID Path
----------- ----------- ----------------------
18 17 0,8,12,16,17,18
19 17 0,8,12,16,17,19
就这么简单，实际上有Sql server 2008中HierarchyType 也能很好的解决这个问题。我将在后面写一些关于HierarchyType的Post.

希望这篇POST对您有帮助。

Author Peter Liu