with cte as
{
    select id,parent_id
    from tb
    where parent_id = 0
    union all
    select e.id,e.parent_id
    from tb e join cte f on e.id = f.parent_id
}

--类似了。

------解决方案--------------------
SQL code
SELECT * FROM 表 where parent_id in 
(select id from 表 where parent_id='0')

------解决方案--------------------
SQL code
--参考一下实例
--> 生成测试数据表:tb
IF NOT OBJECT_ID('[tb]') IS NULL
 DROP TABLE [tb]
GO
CREATE TABLE [tb](GUID INT IDENTITY,[col1] NVARCHAR(10),[col2] NVARCHAR(20))
INSERT [tb]
SELECT N'A','01' UNION ALL
SELECT N'B','01.01' UNION ALL
SELECT N'C','01.01.01' UNION ALL
SELECT N'F','01.01.01.01' UNION ALL
SELECT N'E','01.01.01.02' UNION ALL
SELECT N'D','01.01.01.03' UNION ALL
SELECT N'O','02' UNION ALL
SELECT N'P','02.01' UNION ALL
SELECT N'Q','02.01.01' 
GO
--SELECT * FROM [tb]

-->SQL查询如下:

---另一种方法
;WITH T AS
(
   SELECT *,PATH=CAST([COL1] AS VARCHAR(1000)) FROM TB A
       WHERE NOT EXISTS(
        SELECT 1 FROM TB 
     WHERE A.COL2 LIKE COL2+'%' 
   AND LEN(A.COL2)>LEN(COL2))
   UNION ALL
   SELECT A.*,CAST(PATH+'-->'+A.COL1 AS VARCHAR(1000))
   FROM TB A 
   JOIN T B 
        ON A.COL2 LIKE B.COL2+'%' 

           AND LEN(A.COL2)-3=LEN(B.COL2)
)

SELECT * FROM T ORDER BY LEFT(COL2,2)

/*

GUID        COL1        COL2                  PATH

----------- ---------- -------------------- --------------------

1           A          01                   A

2           B          01.01                A-->B

3           C          01.01.01             A-->B-->C

4           F          01.01.01.01          A-->B-->C-->F

5           E          01.01.01.02          A-->B-->C-->E

6           D          01.01.01.03          A-->B-->C-->D

7           O          02                   O

8           P          02.01                O-->P

9           Q          02.01.01             O-->P-->Q
(9 行受影响)

*/


;WITH T AS

(
    SELECT *,CAST(COL1  AS VARCHAR(1000)) AS PATH
    FROM  TB 
    WHERE COL2 NOT LIKE '%.%'
    UNION ALL
    SELECT A.*,CAST(B.PATH+'-->'+A.COL1 AS VARCHAR(1000))
    FROM TB A,T B
    WHERE A.COL2 LIKE B.COL2+'.[01-99][01-99]'
)

SELECT * FROM T 
ORDER BY LEFT(COL2,2)

/*

GUID        COL1        COL2                  PATH

----------- ---------- -------------------- --------------------

1           A          01                   A

2           B          01.01                A-->B

3           C          01.01.01             A-->B-->C

4           F          01.01.01.01          A-->B-->C-->F

5           E          01.01.01.02          A-->B-->C-->E

6           D          01.01.01.03          A-->B-->C-->D

7           O          02                   O

8           P          02.01                O-->P

9           Q          02.01.01             O-->P-->Q

 (9 行受影响)

*/

------解决方案--------------------
SQL code
/*
标题：SQL SERVER 2000中查询指定节点及其所有父节点的函数(字符串形式显示)
作者：爱新觉罗·毓华(十八年风雨,守得冰山雪莲花开)  
时间：2010-02-02
地点：新疆乌鲁木齐
*/

create table tb(id varchar(3) , pid varchar(3) , name var