合并列值
--*******************************************************************************************
表结构，数据如下：
id    value
----- ------
   aa bb aaa bbb ccc 需要得到结果：
id    values
------ -----------
     aa,bb aaa,bbb,ccc
即：group by id, 求 value 的和（字符串相加） 1. 旧的解决方法(在sql server 2000中只能用函数解决。)
--=============================================================================
create table tb(id int, value varchar(10))
insert into tb values(1, 'aa')
insert into tb values(1, 'bb')
insert into tb values(2, 'aaa')
insert into tb values(2, 'bbb')
insert into tb values(2, 'ccc')
go
--1. 创建处理函数
CREATE FUNCTION dbo.f_strUnite(@id int)
RETURNS varchar(8000)
AS
BEGIN DECLARE @str varchar(8000) SET @str = '' SELECT @str = @str + ',' + value FROM tb WHERE id=@id RETURN STUFF(@str, 1, 1, '')
END
GO
-- 调用函数
SELECt id, value = dbo.f_strUnite(id) FROM tb GROUP BY id
drop table tb
drop function dbo.f_strUnite
go
/*
id          value
----------- ----------- aa,bb aaa,bbb,ccc
（所影响的行数为 2 行）
*/
--===================================================================================
2. 新的解决方法(在sql server 2005中用OUTER APPLY等解决。)
create table tb(id int, value varchar(10))
insert into tb values(1, 'aa')
insert into tb values(1, 'bb')
insert into tb values(2, 'aaa')
insert into tb values(2, 'bbb')
insert into tb values(2, 'ccc')
go
-- 查询处理
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
drop table tb /*
id          values
----------- ----------- aa,bb aaa,bbb,ccc (2 行受影响)
*/ --SQL2005中的方法2
create table tb(id int, value varchar(10))
insert into tb values(1, 'aa')
insert into tb values(1, 'bb')
insert into tb values(2, 'aaa')
insert into tb values(2, 'bbb')
insert into tb values(2, 'ccc')
go select id, [values]=stuff((select ','+[value] from tb t where id=tb.id for xml path('')), 1, 1, '')
from tb
group by id /*
id          values
----------- -------------------- aa,bb aaa,bbb,ccc (2 row(s) affected) */

selectBeginCity,EndCity,FanDian,[CangWei]=stuff((select '/'+[CangWei] from test1 t  where t.BeginCity=Test1.BeginCity  and t.EndCity=Test1.EndCity  and t.FanDian=Test1.FanDian   for xml path('')), 1, 1, '')
from
Test1
group by
BeginCity,EndCity,FanDian