求SQL语句

[bashanhu]

环境：SQL SERVER2000
原始表：CREATE TABLE [kqday] (
[gh] [int] NOT NULL ,
[rq] [char] (2) NOT NULL ,
[sour] [varchar] (60) NOT NULL ,
PRIMARY KEY CLUSTERED
(
[gh],
[rq]
) ON [PRIMARY]
) ON [PRIMARY]
GO
原始表数据：
gh rq sour
4 11 07:53
4 12 07:53,12:04,13:23
4 17 07:58,12:04
5 01 07:53,12:04,13:23,17:35
......

目的表：CREATE TABLE [kqtime] (
[gh] [char] (15) NOT NULL ,
[rq] [char] (2) NOT NULL ,
[sktime] [char] NOT NULL
) ON [PRIMARY]
GO
需要的数据：
gh rq sktime
4 11 07:53
4 12 07:53
4 12 12:04
4 12 13:23
4 17 07:58
4 17 12:04
5 01 07:53
5 01 12:04
5 01 13:23
5 01 17:35

[--小F--]

合并拆分表
[code=SQL]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
    charindex(',',','+a.Col2,b.ID)=b.ID --也可用 substring(','+a.COl2,b.ID,1)=','
--2000不使用辅助表
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
*/

[/code]

[nianran520]

--> 测试数据：[kqday]
if object_id('[kqday]') is not null drop table [kqday]
create table [kqday]([gh] int,[rq] varchar(2),[sour] varchar(23))
insert [kqday]
select 4,'11','07:53' union all
select 4,'12','07:53,12:04,13:23' union all
select 4,'17','07:58,12:04' union all
select 5,'01','07:53,12:04,13:23,17:35'
--> 测试数据：[kqtime]
if object_id('[kqtime]') is not null drop table [kqtime]
CREATE TABLE [kqtime] ( 
[gh] [char] (15) NOT NULL , 
[rq] [char] (2) NOT NULL , 
[sktime] [char](20) NOT NULL 
) ON [PRIMARY] 
GO 
insert into kqtime
select 
[gh],
[rq], 
[sour]=substring(r.[sour],t.number,charindex(',',r.[sour]+',',t.number)-t.number)
from [kqday] r,master..spt_values t
where t.type='p' 
and substring(','+r.[sour],t.number,1)=','

select * from kqtime
----------------------------
4              	11	07:53               
4              	12	07:53               
4              	12	12:04               
4              	12	13:23               
4              	17	07:58               
4              	17	12:04               
5              	01	07:53               
5              	01	12:04               
5              	01	13:23               
5              	01	17:35               

[让你望见影子的墙]

小FMM还在啊

[--小F--]

[Quote=引用 3 楼 herowang 的回复:]
小FMM还在啊
[/Quote]

回去了 无聊又打开了电脑

[ACMAIN_CHM]

小FMM

小肥猫猫？

[dawugui]

--sql 2000需要使用下面的临时表来做.

create table [tb]([gh] int,[rq] varchar(2),[sour] varchar(23))
insert [tb]
select 4,'11','07:53' union all
select 4,'12','07:53,12:04,13:23' union all
select 4,'17','07:58,12:04' union all
select 5,'01','07:53,12:04,13:23,17:35'
go

SELECT TOP 8000 id = IDENTITY(int, 1, 1) INTO # FROM syscolumns a, syscolumns b 

SELECT A.gh , a.rq, sour = SUBSTRING(A.[sour], B.id, CHARINDEX(',', A.[sour] + ',', B.id) - B.id)
FROM tb A, # B
WHERE SUBSTRING(',' + A.[sour], B.id, 1) = ','
order by A.gh , a.rq

DROP TABLE #
drop table tb

/*
gh          rq   sour                    
----------- ---- ----------------------- 
4           11   07:53
4           12   07:53
4           12   12:04
4           12   13:23
4           17   07:58
4           17   12:04
5           01   07:53
5           01   12:04
5           01   13:23
5           01   17:35

（所影响的行数为 10 行）

*/

--其他方法见下:

/*
标题：分拆列值1
作者：爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)
时间：2008-11-20
地点：广东深圳
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)
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 #

--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)
*/

[fellatioyzx]

学习了

[bashanhu]

烦劳各位大侠了！祝CSDN的兄弟姐妹们元旦快乐！