--> 测试数据: #T
if object_id('tempdb.dbo.#T') is not null drop table #T
create table #T (productId varchar(9),keywords varchar(20),productName varchar(11))
insert into #T
select '01','mp3,mp4,mp5','音乐播放器' union all
select '02','video','视频播放器'select top 20 id=identity(int,1,1) into # from syscolumns aselect substring(a.keywords,b.id,charindex(',',a.keywords+',',b.id)-b.id) s
from #T a,# b
where substring(','+a.keywords,b.id,1)=','
if object_id('tempdb.dbo.#T') is not null drop table #T
create table #T (productId varchar(9),keywords varchar(20),productName varchar(11))
insert into #T
select '01','mp3,mp4,mp5','音乐播放器' union all
select '02','video','视频播放器'select top 20 id=identity(int,1,1) into # from syscolumns aselect substring(a.keywords,b.id,charindex(',',a.keywords+',',b.id)-b.id) s
from #T a,# b
where substring(','+a.keywords,b.id,1)=','
if object_id('tempdb.dbo.#T') is not null drop table #T
create table #T (productId varchar(9),keywords varchar(20),productName varchar(11))
insert into #T
select '01','mp3,mp4,mp5','音乐播放器' union all
select '02','video','视频播放器'select top 20 id=identity(int,1,1) into # from syscolumns aselect id=identity(int,1,1),substring(a.keywords,b.id,charindex(',',a.keywords+',',b.id)-b.id) keyword
into #temp
from #T a,# b
where substring(','+a.keywords,b.id,1)=','select * from #tempdrop table #T,#,#temp/*
id keyword
----------- --------------------
1 mp3
2 mp4
3 mp5
4 video(4 行受影响)
*/
2 aaa,bbb,ccc欲按,分拆values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SELECT TOP 8000
id = IDENTITY(int, 1, 1)
INTO #
FROM syscolumns a, syscolumns b
SELECT
A.id,
SUBSTRING(A.[values], B.id, CHARINDEX(',', A.[values] + ',', B.id) - B.id)
FROM tb A, # B
WHERE SUBSTRING(',' + A.[values], B.id, 1) = ','
DROP TABLE #-- 2. 新的解决方法 -- 示例数据DECLARE @t TABLE(id int, [values] varchar(100))
INSERT @t SELECT 1, 'aa,bb'
UNION ALL SELECT 2, 'aaa,bbb,ccc'
-- 查询处理SELECT
A.id, B.value
FROM(
SELECT id, [values] = CONVERT(xml,
'<root><v>' + REPLACE([values], ',', '</v><v>') + '</v></root>')
FROM @t
)A
OUTER APPLY(
SELECT value = N.v.value('.', 'varchar(100)')
FROM A.[values].nodes('/root/v') N(v)
)B
/*--结果id value----------- --------1 aa
1 bb
2 aaa
2 bbb
2 ccc
(5 行受影响)--*/
CSDN 社区帖子地址 附: 合并与分拆的CLR, sql2005的示例中有:
在安装sql 2005的示例后,默认安装目录为 drive:\Program Files\Microsoft SQL Server\90\Samples\Engine\Programmability\CLR\StringUtilities中
在哪儿啊! 如果在深圳, 请你请饭啊.... 呵呵...CSDN怎么搞的, 结贴怎么给的都是0分...