日期:2014-05-18 浏览次数:20739 次
IF OBJECT_ID('tb') IS NOT NULL
    DROP TABLE tb;
GO
CREATE TABLE tb
( id INT identity(1,1),
  vt VARCHAR(500)
);
GO
INSERT INTO tb(vt) VALUES('1,2ab,c3,4eft');
INSERT INTO tb(vt) VALUES('1a,3,B');
select * from tb;
--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 
(5 行受影响) 
*/
------解决方案--------------------
-- 用with语句
;with t as(
    select 
        id, 
        vt=(case when charindex(',',vt)>0 then substring(vt, charindex(',',vt)+1, len(vt)) else '' end), 
        st=(case when charindex(',',vt)>0 then left(vt, charindex(',',vt)-1) else vt end) 
    from tb
    union all
    select 
        t.id, 
        vt=(case when charindex(',',t.vt)>0 then substring(t.vt, charindex(',',t.vt)+1, len(t.vt)) else '' end), 
        st=(case when charindex(',',t.vt)>0 then left(t.vt, charindex(',',t.vt)-1) else t.vt end) 
    from t
    where len(t.vt)>0
)
select id, st from t
order by id