例如：查询结果为
- SQL code
MemberID Name Sex Email Interest000001 张三 男 [email protected] 打球000001 张三 男 [email protected] 游泳000001 张三 男 [email protected] 看电影
我想把这些记录何为一条,如：
- SQL code
MemberID Name Sex Email Interest000001 张三 男 [email protected] 打球、游泳、看电影
怎么办？
------解决方案--------------------
- SQL code
if object_id('tb') is not null drop table tbgocreate table tb( MemberID varchar(10), Name varchar(10), Sex varchar(10), Email varchar(20), Interest varchar(10))goinsert into tbselect '000001','张三','男',[email protected]','打球' union allselect '000001','张三','男',[email protected]','游泳' union allselect '000001','张三','男',[email protected]','看电影'goselect MemberID,Name=Max(Name),Sex=Max(Sex),Email=Max(Email), Interest=stuff((select ','+Interest from tb where MemberID=a.MemberID for xml path('')),1,1,'') from tb a group by MemberIDgo/*MemberID Name Sex Email Interest---------- ---------- ---------- -------------------- ----------------------------------------------------------------------------------------------------------------000001 张三 男 [email protected] 打球,游泳,看电影(1 行受影响)*/
------解决方案--------------------
- SQL code
create table xp447196763(MemberID varchar(8), name varchar(4), Sex varchar(4), Email varchar(15), Interest varchar(8))insert into xp447196763select '000001', '张三', '男', [email protected]', '打球' union allselect '000001', '张三', '男', [email protected]', '游泳' union allselect '000001', '张三', '男', [email protected]', '看电影'declare @Interests varchar(50)=''select @[email protected]+Interest+'、'from xp447196763select MemberID,name,Sex,Email,left(@Interests,len(@Interests)-1) Interestfrom xp447196763group by MemberID,name,Sex,EmailMemberID name Sex Email Interest-------- ---- ---- --------------- --------------------------------------------------000001 张三 男 [email protected] 打球、游泳、看电影(1 row(s) affected)
------解决方案--------------------
- SQL code
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 [email protected] 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) */