拆分吧?/* 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) */
---------------------------------------------------------------- -- Author :fredrickhu(小F,向高手学习) -- Date :2011-11-14 17:07:38 -- Version: -- Microsoft SQL Server 2008 R2 (RTM) - 10.50.1617.0 (Intel X86) -- Apr 22 2011 11:57:00 -- Copyright (c) Microsoft Corporation -- Enterprise Evaluation Edition on Windows NT 6.1 <X64> (Build 7600: ) (WOW64) -- ---------------------------------------------------------------- --> 测试数据:[tb] if object_id('[tb]') is not null drop table [tb] go create table [tb]([uid] varchar(1),[list] varchar(7),[tid] varchar(2)) insert [tb] select 'A','|1|8|2|','00' union all select 'B','|1|3|2|','tt' union all select 'A','|1|9|4|','ii' --------------开始查询-------------------------- Select a.uid,COl2=substring(a.list,b.number,charindex('|',a.list+'|',b.number)-b.number),tid from Tb a join master..spt_values b ON B.type='p' AND B.number BETWEEN 1 AND LEN(A.list) where substring('|'+a.list,b.number,1)='|' ----------------结果---------------------------- /* uid COl2 tid ---- ------- ---- A 00 A 1 00 A 8 00 A 2 00 B tt B 1 tt B 3 tt B 2 tt A ii A 1 ii A 9 ii A 4 ii(12 行受影响)*/
create table tb(uid char(1),list varchar(20),tid varchar(10)) insert into tb select 'A','|1|8|2|','oo' union select 'B','|1|3|2|','tt' union select 'A','|1|9|4|','ii'select a.uid, list=substring(a.list,b.number,charindex('|',a.list,b.number)-b.number), a.tid from tb a,master..spt_values b where b.[type] = 'p' and b.number between 2 and len(a.list) and substring('|'+a.list,b.number,1) = '|'/* uid list tid ---- -------------------- ---------- A 1 oo A 8 oo A 2 oo A 1 ii A 9 ii A 4 ii B 1 tt B 3 tt B 2 tt(9 行受影响)
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)
*/
-- Author :fredrickhu(小F,向高手学习)
-- Date :2011-11-14 17:07:38
-- Version:
-- Microsoft SQL Server 2008 R2 (RTM) - 10.50.1617.0 (Intel X86)
-- Apr 22 2011 11:57:00
-- Copyright (c) Microsoft Corporation
-- Enterprise Evaluation Edition on Windows NT 6.1 <X64> (Build 7600: ) (WOW64)
--
----------------------------------------------------------------
--> 测试数据:[tb]
if object_id('[tb]') is not null drop table [tb]
go
create table [tb]([uid] varchar(1),[list] varchar(7),[tid] varchar(2))
insert [tb]
select 'A','|1|8|2|','00' union all
select 'B','|1|3|2|','tt' union all
select 'A','|1|9|4|','ii'
--------------开始查询--------------------------
Select
a.uid,COl2=substring(a.list,b.number,charindex('|',a.list+'|',b.number)-b.number),tid
from
Tb a join master..spt_values b
ON B.type='p' AND B.number BETWEEN 1 AND LEN(A.list)
where
substring('|'+a.list,b.number,1)='|'
----------------结果----------------------------
/* uid COl2 tid
---- ------- ----
A 00
A 1 00
A 8 00
A 2 00
B tt
B 1 tt
B 3 tt
B 2 tt
A ii
A 1 ii
A 9 ii
A 4 ii(12 行受影响)*/
insert into tb
select 'A','|1|8|2|','oo' union
select 'B','|1|3|2|','tt' union
select 'A','|1|9|4|','ii'select a.uid,
list=substring(a.list,b.number,charindex('|',a.list,b.number)-b.number),
a.tid
from tb a,master..spt_values b
where b.[type] = 'p' and b.number between 2 and len(a.list)
and substring('|'+a.list,b.number,1) = '|'/*
uid list tid
---- -------------------- ----------
A 1 oo
A 8 oo
A 2 oo
A 1 ii
A 9 ii
A 4 ii
B 1 tt
B 3 tt
B 2 tt(9 行受影响)