修正一个BUG
原帖出处 : http://bbs.csdn.net/topics/270018720
分拆列值
原著:邹建
改编:爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开) 2007-12-16 广东深圳
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, 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(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
------解决方案--------------------
&的问题,替换掉就ok了。
if object_id('[tb]') is not null drop table [tb]
go
create table tb(id int,value varchar(30))
insert into tb values(1,'a&a,b&b')
go
declare @sign varchar(36) set @sign=newid()
declare @t table(id int,value varchar(400))
<