参考:/* 标题:简单数据拆分(version 2.0) 作者:爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开) 时间:2010-05-07 地点:重庆航天职业学院 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)create table tb(id int,value varchar(30)) insert into tb values(1,'aa,bb') insert into tb values(2,'aaa,bbb,ccc') go--方法1.使用临时表完成 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 #--方法2.如果数据量小,可不使用临时表 select a.id , value = substring(a.value , b.number , charindex(',' , a.value + ',' , b.number) - b.number) from tb a join master..spt_values b on b.type='p' and b.number between 1 and len(a.value) where substring(',' + a.value , b.number , 1) = ','--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--方法1.使用xml完成 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--方法2.使用CTE完成 ;with tt as (select id,[value]=cast(left([value],charindex(',',[value]+',')-1) as nvarchar(100)),Split=cast(stuff([value]+',',1,charindex(',',[value]+','),'') as nvarchar(100)) from tb union all select id,[value]=cast(left(Split,charindex(',',Split)-1) as nvarchar(100)),Split= cast(stuff(Split,1,charindex(',',Split),'') as nvarchar(100)) from tt where split>'' ) select id,[value] from tt order by id option (MAXRECURSION 0) DROP TABLE tb/* id value ----------- ------------------------------ 1 aa 1 bb 2 aaa 2 bbb 2 ccc(5 行受影响) */
select left(a , charindex(',',a) - 1) a from ( select a = substring(a.a , b.number , charindex('|' , a.a + '|' , b.number) - b.number) from tb a join master..spt_values b on b.type='p' and b.number between 1 and len(a.a) where substring('|' + a.a , b.number , 1) = '|' ) t
if object_id('tb') is not null drop table tb go create table tb ( a varchar(60) ) insert into tb select '10号车,10|11号车,11' union all select '22机动车,22|23柴油车,23|25AA,25|26,26' go select substring(substring(a,number,charindex('|',a+'|',number)-number),1,charindex(',',substring(a,number,charindex('|',a+'|',number)-number))-1) from tb cross join master..spt_values where type='p' and number between 1 and len(a) and substring('|'+a,number,1)='|' go /*------------------------------------------------------------ 10号车 11号车 22机动车 23柴油车 25AA 26(6 行受影响)*/
select substring(substring(a,number,charindex('|',a+'|',number)-number),1,charindex(',',substring(a,number,charindex('|',a+'|',number)-number))-1) from tb , master..spt_values where type='p' and number between 1 and len(a) and substring('|'+a,number,1)='|'
with t(p1,p2,name) as ( select charindex('|','|'+name),charindex('|',name+'|')+1,name from [tb] union all select b.p2,charindex('|',a.name+'|',b.p2)+1,a.name from [tb] a join t b on a.name=b.name where charindex('|',a.name+'|',b.p2)>0 ) select name=left(substring(name,p1,p2-p1-1),charindex(',',substring(name,p1,p2-p1-1))-1) from t order by name,p1
标题:简单数据拆分(version 2.0)
作者:爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)
时间:2010-05-07
地点:重庆航天职业学院
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)create table tb(id int,value varchar(30))
insert into tb values(1,'aa,bb')
insert into tb values(2,'aaa,bbb,ccc')
go--方法1.使用临时表完成
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 #--方法2.如果数据量小,可不使用临时表
select a.id , value = substring(a.value , b.number , charindex(',' , a.value + ',' , b.number) - b.number)
from tb a join master..spt_values b
on b.type='p' and b.number between 1 and len(a.value)
where substring(',' + a.value , b.number , 1) = ','--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--方法1.使用xml完成
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--方法2.使用CTE完成
;with tt as
(select id,[value]=cast(left([value],charindex(',',[value]+',')-1) as nvarchar(100)),Split=cast(stuff([value]+',',1,charindex(',',[value]+','),'') as nvarchar(100)) from tb
union all
select id,[value]=cast(left(Split,charindex(',',Split)-1) as nvarchar(100)),Split= cast(stuff(Split,1,charindex(',',Split),'') as nvarchar(100)) from tt where split>''
)
select id,[value] from tt order by id option (MAXRECURSION 0)
DROP TABLE tb/*
id value
----------- ------------------------------
1 aa
1 bb
2 aaa
2 bbb
2 ccc(5 行受影响)
*/
(
select a = substring(a.a , b.number , charindex('|' , a.a + '|' , b.number) - b.number)
from tb a join master..spt_values b
on b.type='p' and b.number between 1 and len(a.a)
where substring('|' + a.a , b.number , 1) = '|'
) t
if object_id('tb') is not null
drop table tb
go
create table tb
(
a varchar(60)
)
insert into tb
select '10号车,10|11号车,11' union all
select '22机动车,22|23柴油车,23|25AA,25|26,26'
go
select substring(substring(a,number,charindex('|',a+'|',number)-number),1,charindex(',',substring(a,number,charindex('|',a+'|',number)-number))-1) from tb cross join master..spt_values where type='p' and number between 1 and len(a) and substring('|'+a,number,1)='|'
go
/*------------------------------------------------------------
10号车
11号车
22机动车
23柴油车
25AA
26(6 行受影响)*/
substring(substring(a,number,charindex('|',a+'|',number)-number),1,charindex(',',substring(a,number,charindex('|',a+'|',number)-number))-1)
from
tb , master..spt_values
where
type='p' and number between 1 and len(a) and substring('|'+a,number,1)='|'
(
select charindex('|','|'+name),charindex('|',name+'|')+1,name from [tb]
union all
select b.p2,charindex('|',a.name+'|',b.p2)+1,a.name
from [tb] a join t b
on a.name=b.name
where charindex('|',a.name+'|',b.p2)>0
)
select name=left(substring(name,p1,p2-p1-1),charindex(',',substring(name,p1,p2-p1-1))-1)
from t order by name,p1