字符串连接函数的BUG? VARCHAR(MAX)

create table # (a int, b int, c varchar(8000), create_date datetime)



insert # select 1,1, 'a', 0

insert # select 1,1, 'b', 1



insert # select 1,2, 'c', 2

insert # select 1,2, 'd', 3





declare @ varchar(max)

select @ = isnull(@+'_', '') + ltrim(c) from # where a=1 and b=1 order by create_date desc

select @ -- b_a



/*

这里没有问题

你看看函数的执行计划，要不把缓存清一下试试

*/

/*

标题：按某字段合并字符串之一(简单合并)

作者：爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)

时间：2008-11-06

地点：广东深圳



描述：将如下形式的数据按id字段合并value字段。

id    value

----- ------

1     aa

1     bb

2     aaa

2     bbb

2     ccc

需要得到结果：

id     value

------ -----------

1      aa,bb

2      aaa,bbb,ccc

即：group by id, 求 value 的和（字符串相加）

*/

--1、sql2000中只能用自定义的函数解决

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



create function dbo.f_str(@id varchar(10)) returns varchar(1000)

as

begin

  declare @str varchar(1000)

  select @str = isnull(@str + ',' , '') + cast(value as varchar) from tb where id = @id

  return @str

end

go



--调用函数

select id , value = dbo.f_str(id) from tb group by id



drop function dbo.f_str

drop table tb





--2、sql2005中的方法

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, [value] = stuff((select ',' + [value] from tb t where id = tb.id for xml path('')) , 1 , 1 , '')

from tb

group by id



drop table tb





--3、使用游标合并数据

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

declare @t table(id int,value varchar(100))--定义结果集表变量

--定义游标并进行合并处理

declare my_cursor cursor local for

select id , value from tb

declare @id_old int , @id int , @value varchar(10) , @s varchar(100)

open my_cursor

fetch my_cursor into @id , @value

select @id_old = @id , @s=''

while @@FETCH_STATUS = 0

begin

    if @id = @id_old

       select @s = @s + ',' + cast(@value as varchar)

    else

      begin

        insert @t values(@id_old , stuff(@s,1,1,''))

        select @s = ',' + cast(@value as varchar) , @id_old = @id

      end

    fetch my_cursor into @id , @value

END

insert @t values(@id_old , stuff(@s,1,1,''))

close my_cursor

deallocate my_cursor



select * from @t

drop table tb

最好给出完整的表结构,测试数据,计算方法和正确结果.否则耽搁的是你宝贵的时间。

如果有多表,表之间如何关联?

declare @s  varchar(8000) 

select @s=isnull(@s+'_','')+ltrim(c)

from (select 2 as c union all select 3) temp

select @s



go



declare @s  varchar(max) 

select @s=isnull(@s+'_','')+ltrim(c)

from (select 2 as c union all select 3) temp

select @s





你想证明什么呢？

--贴出你的数据和结果，顺便拿来ID、IC...卡和密码。

alter function [dbo].[f_str](@a varchar(20),@b varchar(20))

returns varchar(MAX)

as

begin

	return stuff((select '_'+ltrim(c) from tb where a=@a and b=@b order by tb.created desc for xml path('')), 1, 1, '')

end