SQL2000 函数递归显示路径

----------------------------------------------------------------------------------
-- Author : htl258(Tony)
-- Date : 2011-05-19 17:25:59
-- Version: Microsoft SQL Server 2008 (RTM) - 10.0.1600.22 (Intel X86)
-- Jul 9 2008 14:43:34
-- Copyright (c) 1988-2008 Microsoft Corporation
-- Developer Edition on Windows NT 5.1 <X86> (Build 2600: Service Pack 2)
-- Blog : http://blog.csdn.net/htl258 ----------------------------------------------------------------------------------
--> 生成测试数据表: [Tree]
IF OBJECT_ID('[Tree]') IS NOT NULL
DROP TABLE [Tree]
GO
CREATE TABLE [Tree] ([ID] [int],[Text] [nvarchar](10),[PID] [int])
INSERT INTO [Tree]
SELECT '1','A',NULL UNION ALL
SELECT '2','B','1' UNION ALL
SELECT '3','C','2' UNION ALL
SELECT '4','D','3'

--SELECT * FROM [Tree]
GO
-->SQL查询如下:

IF OBJECT_ID('getdpt')>0
DROP FUNCTION getdpt
GO
CREATE FUNCTION GetDPT(@Id INT)
RETURNS NVARCHAR(200)
AS
BEGIN
DECLARE @dptn NVARCHAR(50), @dpti NVARCHAR(20)
SELECT @dptn = [Text], @dpti = PID
FROM Tree
WHERE ID = @Id
RETURN
CASE
WHEN ISNULL(@dpti, '')='' THEN @dptn
ELSE ISNULL(dbo.GetDPT(@dpti)+'->', '')+@dptn
END
END
GO

SELECT dbo.GetDPT(4)
/*
A->B->C->D

(1 行受影响)
*/