表A:
name num
aa 5/6/9
bb 6/7/8想要的结果:
col1 col2
5 aa
6 aa
6 bb
7 bb
8 bb
9 aa哪位知道应该如何查询
name num
aa 5/6/9
bb 6/7/8想要的结果:
col1 col2
5 aa
6 aa
6 bb
7 bb
8 bb
9 aa哪位知道应该如何查询
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--拆分表:--> --> (Roy)生成測試數據
if not object_id('Tab') is null
drop table Tab
Go
Create table Tab([Col1] int,[COl2] nvarchar(5))
Insert Tab
select 1,N'a,b,c' union all
select 2,N'd,e' union all
select 3,N'f'
Go--SQL2000用辅助表:
if object_id('Tempdb..#Num') is not null
drop table #Num
go
select top 100 ID=Identity(int,1,1) into #Num from syscolumns a,syscolumns b
Select
a.Col1,COl2=substring(a.Col2,b.ID,charindex(',',a.Col2+',',b.ID)-b.ID)
from
Tab a,#Num b
where
substring(','+a.COl2,b.ID,1)=','
--2000不适用辅助表
Select
a.Col1,COl2=substring(a.Col2,b.number,charindex(',',a.Col2+',',b.number)-b.number)
from
Tab a join master..spt_values b
ON B.type='p' AND B.number BETWEEN 1 AND LEN(A.col2)
where
substring(','+a.COl2,b.number,1)=','
--SQL2005用Xml:select
a.COl1,b.Col2
from
(select Col1,COl2=convert(xml,'<root><v>'+replace(COl2,',','</v><v>')+'</v></root>') from Tab)a
outer apply
(select Col2=C.v.value('.','nvarchar(100)') from a.COl2.nodes('/root/v')C(v))b
--SQL05用CTE:;with roy as
(select Col1,COl2=cast(left(Col2,charindex(',',Col2+',')-1) as nvarchar(100)),Split=cast(stuff(COl2+',',1,charindex(',',Col2+','),'') as nvarchar(100)) from Tab
union all
select Col1,COl2=cast(left(Split,charindex(',',Split)-1) as nvarchar(100)),Split= cast(stuff(Split,1,charindex(',',Split),'') as nvarchar(100)) from Roy where split>''
)
select COl1,COl2 from roy order by COl1 option (MAXRECURSION 0)--生成结果:
/*
Col1 COl2
----------- -----
1 a
1 b
1 c
2 d
2 e
3 f
*/
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)
*/
create table test(name varchar(20),num varchar(20))
insert test
select 'aa','5/6/9' union all
select 'bb','6/7/8'
go--drop table testselect
a.name[col2],col1=substring(a.num,b.number,charindex('/',a.num+'/',b.number)-b.number)
from
test a join master..spt_values b
ON B.type='p' AND B.number BETWEEN 1 AND LEN(A.num)
where
substring('/'+a.num,b.number,1)='/'
order by col1 col2 col1
-------------------- --------------------
aa 5
aa 6
bb 6
bb 7
bb 8
aa 9(所影响的行数为 6 行)
-- Author :fredrickhu(我是小F,向高手学习)
-- Date :2009-09-08 16:19:47
-- Version:
-- Microsoft SQL Server 2005 - 9.00.4035.00 (Intel X86)
-- Nov 24 2008 13:01:59
-- Copyright (c) 1988-2005 Microsoft Corporation
-- Developer Edition on Windows NT 5.2 (Build 3790: Service Pack 1)
--
----------------------------------------------------------------
--> 测试数据:[tb]
if object_id('[tb]') is not null drop table [tb]
go
create table [tb]([name] varchar(2),[num] varchar(20))
insert [tb]
select 'aa','5/6/9' union all
select 'bb','6/7/8'
--------------开始查询--------------------------
SELECT A.name, B.num
FROM(
SELECT name, [num] = CONVERT(xml,' <root> <v>' + REPLACE([num], '/', ' </v> <v>') + ' </v> </root>') FROM tb
)A
OUTER APPLY(
SELECT num = N.v.value('.', 'varchar(100)') FROM A.[num].nodes('/root/v') N(v)
)B ----------------结果----------------------------
/* name num
---- ----------------------------------------------------------------------------------------------------
aa 5
aa 6
aa 9
bb 6
bb 7
bb 8 (6 行受影响)*/