日期:2014-05-18  浏览次数:20647 次

行列转换的动态sql
数据:
 id customer_guid customer_name
 01 1101 李炳
 02 1001 赵田
 02 1002 肃静
 03 2101 赵翼
 03 2102 李田
 03 2103 苏静


sql后的结果
  
  id customer_name
  01 李炳
  02 赵田,肃静
  03 赵翼,李田,苏静

说明: ,是必须的 而且具体的customer_guid是动态的,有可能有更多

请大家指点,希望能给用动态 sql给出答案。

------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 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) 

*/ 

drop table tb

------解决方案--------------------
SQL code
--> --> (Roy)生成測試數據
 
if not object_id('Tempdb..#T') is null
    drop table #T
Go
Create table #T([id] nvarchar(2),[customer_guid] int,[customer_name] nvarchar(2))
Insert #T
select N'01',1101,N'李炳' union all
select N'02',1001,N'赵田' union all
select N'02',1002,N'肃静' union all
select N'03',2101,N'赵翼' union all
select N'03',2102,N'李田' union all
select N'03',2103,N'苏静'
Go
select 
    a.[id],[customer_name]=stuff(b.[customer_name].value('/R[1]','nvarchar(max)'),1,1,'')
from 
    (select distinct [id] from #T) a
Cross apply
    (select [customer_name]=(select N','+[customer_name] from #T where [id]=a.[id] For XML PATH(''), ROOT('R'), TYPE))b
    
/*
    id    customer_name
01    李炳
02    赵田,肃静
03    赵翼,李田,苏静
*/