整理的方法如下：
-- =============================================================================
-- Title: 在SQL中分类合并数据行
-- Author: dobear Mail(MSN): [email protected]
-- Environment: Vista + SQL2005
-- Date: 2008-04-22
-- =============================================================================

--1. 创建表，添加测试数据
CREATE TABLE tb(id int, [value] varchar(10))
INSERT tb SELECT 1, 'aa'
UNION ALL SELECT 1, 'bb'
UNION ALL SELECT 2, 'aaa'
UNION ALL SELECT 2, 'bbb'
UNION ALL SELECT 2, 'ccc'

--SELECT * FROM tb
/**//*
id value
----------- ----------
1 aa
1 bb
2 aaa
2 bbb
2 ccc

(5 row(s) affected)
*/


--2 在SQL2000只能用自定义函数实现
----2.1 创建合并函数fn_strSum，根据id合并value值
GO
CREATE FUNCTION dbo.fn_strSum(@id int)
RETURNS varchar(8000)
AS
BEGIN
DECLARE @values varchar(8000)
SET @values = ''
SELECT @values = @values + ',' + value FROM tb WHERE id=@id
RETURN STUFF(@values, 1, 1, '')
END
GO

-- 调用函数
SELECT id, VALUE = dbo.fn_strSum(id) FROM tb GROUP BY id
DROP FUNCTION dbo.fn_strSum

----2.2 创建合并函数fn_strSum2，根据id合并value值
GO
CREATE FUNCTION dbo.fn_strSum2(@id int)
RETURNS varchar(8000)
AS
BEGIN
DECLARE @values varchar(8000)
SELECT @values = isnull(@values + ',', '') + value FROM tb WHERE id=@id
RETURN @values
END
GO

-- 调用函数
SELECT id, VALUE = dbo.fn_strSum2(id) FROM tb GROUP BY id
DROP FUNCTION dbo.fn_strSum2


--3 在SQL2005中的新解法
----3.1 使用OUTER APPLY
SELECT *
FROM (SELECT DISTINCT id FROM tb) A OUTER APPLY(
SELECT [values]= STUFF(REPLACE(REPLACE(
(
SELECT value FROM tb N
WHERE id = A.id
FOR XML AUTO
), '<N value="', ','), '"/>', ''), 1, 1, '')
)N

----3.2 使用XML
SELECT id, [values]=STUFF((SELECT ','+[value] FROM tb t WHERE id=tb.id FOR XML PATH('')), 1, 1, '')
FROM tb
GROUP BY id

--4 删除测试表tb
drop table tb

/**//*
id values
----------- --------------------
1 aa,bb
2 aaa,bbb,ccc

(2 row(s) affected)
*/


roy_88 兄弟整理的方法：

--合并分拆表
/******************************************************************************************************************************************************
合并分拆表数据

整理人：中国风(Roy)

日期:2008.06.06
******************************************************************************************************************************************************/

--> --> (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(1))
Insert Tab
select 1,N'a' union all
select 1,N'b' union all
select 1,N'c' union all
select 2,N'd' union all
select 2,N'e' union all
select 3,N'f'
Go

合并表:

SQL2000用函数:

go
if object_id('F_Str') is not null
drop function F_Str
go
create function F_Str(@Col1 int)
returns nvarchar(100)
as
begin
declare @S nvarchar(100)
select @S=isnull(@S+',','')+Col2 from Tab where Col1=@Col1
return @S
end
go
Select distinct Col1,Col2=dbo.F_Str(Col1) from Tab

go

SQL2005用XML:

方法1:

select
a.Col1,Col2=stuff(b.Col2.value('/R[1]','nvarchar(max)'),1,1,'')
from
(select distinct COl1 from Tab) a
Cross apply
(select COl2=(select N','+Col2 from Tab where Col1=a.COl1 For XML PATH(''), ROOT('R'), TYPE))b

方法2:

select
a.Col1,COl2=replace(b.Col2.value('/Tab[1]','nvarchar(max)'),char(44)+char(32),char(44))
from
(select distinct COl1 from Tab) a
cross apply
(select Col2=(select COl2 from Tab where COl1=a.COl1 FOR XML AUTO, TYPE)
.query('<Tab>
{for $i in /Tab[position()<last()]/@COl2 return concat(string($i),",")}
{concat("",string(/Tab[last()]/@COl2))}
</Tab>')
)b

SQL05用CTE:

;with roy as(select Col1,Col2,row=row_number()over(partition by COl1 order by COl1) from Tab)
,Roy2 as
(select COl1,cast(COl2 as nvarchar(100))COl2,row from Roy where row=1
union all
select a.Col1,cast(b.COl2+','+a.COl2 as nvarchar(100)),a.row from Roy a join Roy2 b _disibledevent=a.COl1) order by Col1 option (MAXRECURSION 0)


生成结果:
/*
Col1 COl2
----------- ------------
1 a,b,c
2 d,e
3 f

(3 行受影响)

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