SQL查询语句 group by后， 字符串合并合并列值 --*******************************************************************************************表结构，数据如下： id value ----- ------ 1 aa 1 bb 2 aaa 2 bbb 2 ccc 需要得到结果： id values ------ ----------- 1 aa,bb 2 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 ----------- ----------- 1 aa,bb 2 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 ----------- ----------- 1 aa,bb 2 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 ----------- -------------------- 1 aa,bb 2 aaa,bbb,ccc (2 row(s) affected) */
demo:
select BeginCity,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
参考: http://bbs.csdn.net/topics/330188225
http://bbs.csdn.net/topics/330182340