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。) --1. 创建处理函数 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
CREATE FUNCTION dbo.f_str(@id int) RETURNS varchar(8000) AS BEGIN DECLARE @r varchar(8000) SET @r = '' SELECT @r = @r + ',' + value FROM tb WHERE id=@id RETURN STUFF(@r, 1, 1, '') END GO
-- 调用函数 SELECt id, value = dbo.f_str(id) FROM tb GROUP BY id
drop table tb drop function dbo.f_str
/* id value ----------- ----------- 1 aa,bb 2 aaa,bbb,ccc （所影响的行数为 2 行） */
--2、另外一种函数. 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
--创建一个合并的函数 create function f_hb(@id int) returns varchar(8000) as begin declare @str varchar(8000) set @str = '' select @str = @str + ',' + cast(value as varchar) from tb where id = @id set @str = right(@str , len(@str) - 1) return(@str) End go
--调用自定义函数得到结果： select distinct id ,dbo.f_hb(id) as value from tb
drop table tb drop function dbo.f_hb
/* 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)
*/
drop table tb拆分列值有表tb, 如下: id value ----------- ----------- 1 aa,bb 2 aaa,bbb,ccc 欲按id,分拆value列, 分拆后结果如下: id value ----------- -------- 1 aa 1 bb 2 aaa 2 bbb 2 ccc
1. 旧的解决方法(sql server 2000) SELECT TOP 8000 id = IDENTITY(int, 1, 1) INTO # FROM syscolumns a, syscolumns b
SELECT A.id, SUBSTRING(A.[values], B.id, CHARINDEX(',', A.[values] + ',', B.id) - B.id) FROM tb A, # B WHERE SUBSTRING(',' + A.[values], B.id, 1) = ','
DROP TABLE #
2. 新的解决方法(sql server 2005)
create table tb(id int,value varchar(30)) insert into tb values(1,'aa,bb') insert into tb values(2,'aaa,bbb,ccc') go SELECT A.id, B.value FROM( SELECT id, [value] = CONVERT(xml,'<root><v>' + REPLACE([value], ',', '</v><v>') + '</v></root>') FROM tb )A OUTER APPLY( SELECT value = N.v.value('.', 'varchar(100)') FROM A.[value].nodes('/root/v') N(v) )B
DROP TABLE tb
/* id value ----------- ------------------------------ 1 aa 1 bb 2 aaa 2 bbb 2 ccc
(5 行受影响) */
转自：http://www.cnblogs.com/hanmos/archive/2011/09/15/2177391.html