--1. 創(chuàng)建表，添加測(cè)試數(shù)據(jù)
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只能用自定義函數(shù)實(shí)現(xiàn)
----2.1 創(chuàng)建合并函數(shù)fn_strSum，根據(jù)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

-- 調(diào)用函數(shù)
SELECT id, VALUE = dbo.fn_strSum(id) FROM tb GROUP BY id
DROP FUNCTION dbo.fn_strSum

----2.2 創(chuàng)建合并函數(shù)fn_strSum2，根據(jù)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

-- 調(diào)用函數(shù)
SELECT id, VALUE = dbo.fn_strSum2(id) FROM tb GROUP BY id
DROP FUNCTION dbo.fn_strSum2


--3 在SQL2005/SQL2008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 刪除測(cè)試表tb
drop table tb

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

(2 row(s) affected)
*/