日期:2014-05-18 浏览次数:20691 次
/*
标题:分拆列值1
作者:爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)
时间:2008-11-20
地点:广东深圳
描述
有表tb, 如下:
id value
----------- -----------
1 aa,bb
2 aaa,bbb,ccc
欲按id,分拆value列, 分拆后结果如下:
id value
----------- --------
1 aa
1 bb
2 aaa
2 bbb
2 ccc
*/
--1. 旧的解决方法(sql server 2000)
SELECT TOP 8000 id = IDENTITY(int, 1, 1) INTO # FROM syscolumns a, syscolumns b
SELECT A.id, value = SUBSTRING(A.[value], B.id, CHARINDEX(',', A.[value] + ',', B.id) - B.id)
FROM tb A, # B
WHERE SUBSTRING(',' + A.[value], B.id, 1) = ','
DROP TABLE #
--2. 新的解决方法(sql server 2005)
create table tb(id int,value varchar(30))
insert into tb values(1,'aa,bb')
insert into tb values(2,'aaa,bbb,ccc')
go
SELECT A.id, B.value
FROM(
SELECT id, [value] = CONVERT(xml,'<root><v>' + REPLACE([value], ',', '</v><v>') + '</v></root>') FROM tb
)A
OUTER APPLY(
SELECT value = N.v.value('.', 'varchar(100)') FROM A.[value].nodes('/root/v') N(v)
)B
DROP TABLE tb
/*
id value
----------- ------------------------------
1 aa
1 bb
2 aaa
2 bbb
2 ccc
(5 行受影响)
*/
------解决方案--------------------
合并分拆表_整理贴1
http://topic.csdn.net/u/20080612/22/c850499f-bce3-4877-82d5-af2357857872.html
------解决方案--------------------
if object_id('t1') is not null
drop table t1
Go
Create table t1([SIC] nvarchar(3),[AC] nvarchar(5),[DESCRIP] nvarchar(10),[OWNERS] nvarchar(20))
Insert into t1
Select N'ECP',N'CP_NC',N'Checkpoint',N'355518,416371'
Union all Select N'ECP',N'CPQ',N'Quickcheck',N'355518,457171,325442'
IF object_id('tempdb..#tmp') IS NOT NULL
DROP TABLE #tmp
SELECT SIC,AC,DESCRIP,OWNERS+',' AS OWNERS INTO #tmp FROM t1
WHILE EXISTS(SELECT 1 FROM #tmp WHERE CHARINDEX(',',OWNERS)>0)
BEGIN
INSERT INTO #tmp(SIC,AC,DESCRIP,OWNERS)
SELECT
SIC,AC,DESCRIP,SUBSTRING(OWNERS,1,CHARINDEX(',',OWNERS)-1)
FROM #tmp
WHERE CHARINDEX(',',OWNERS)>0
UPDATE #tmp
SET OWNERS=STUFF(OWNERS,1,CHARINDEX(',',OWNERS),'')
WHERE CHARINDEX(',',OWNERS)>0
END
SELECT * FROM #tmp WHERE OWNERS>'' ORDER BY SIC,AC,DESCRIP,OWNERS
DROP TABLE #tmp