表結構,資料如下:
id value
----- ------
1 aa
1 bb
2 aaa
2 bbb
2 ccc
需要得到結果:
id values
------ -----------
1 aa,bb
2 aaa,bbb,ccc
即:group by id, 求 value 的和(字串相加)
1. 舊的解決方案(在sql server 2000中只能用函數解決。)
--1. 建立處理函數
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
-- 調用函數
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
(所影響的行數為 2 行)
*/
--2、另外一種函數.
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
--建立一個合并的函數
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
--調用自訂函數得到結果:
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
(所影響的行數為 2 行)
*/