怎么把数据合成一行

[liuyan351677394]

invite_id invite_code plan_code
1141 GZJQ2011001 GZJQ0002
1141 GZJQ2011001 GZJQ0002
1141 GZJQ2011001 GZJQ0001
1121 GZJQ2010003 GZJQ0002
1121 GZJQ2010003 GZJQ0002
1121 GZJQ2010003 GZJQ0001
1112 GZJQ2010002 GZJQ0002
1090 GZJQ2010001 GZJQ0001

把结果合成

invite_id invite_code plan_code
1141 GZJQ2011001 GZJQ0002
GZJQ0001
GZJQ0002

[Mr_Nice]

参考

合并列值 
原著：邹建 
改编：爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)  2007-12-16  广东深圳 

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 

 
 
 create table [user](
id int,name varchar(20)
)
go 
insert [user] select
1,      '张三'
union all select 
2,     '李四' 
union all select 
3,      '王五' 
go
create table skill (
id int,   uid int,   name varchar(20),   level varchar(20)
)
insert skill select
1,      1,      '英语',    '6级' 
union all select 
2,      3,      '日语',    '初级' 
union all select 
3,      1,      '法语',    '入门'
union all select 
4,      3,      '英语',    '4级' 
union all select 
5,      3,      '德语',    '精通' 

go  
  
create function fn_skill(
@uid int,
@name varchar(20)
)
returns varchar(500)
as
begin
   declare @r varchar(500)
   if @name = ''             ---空表示查所有
       select @r = isnull(@r + '、','') + name from skill where uid=@uid
   else
       select @r = isnull(@r + '、','') + name from skill where uid=@uid and name = @name
   return @r
end
go

create function fn_skilldetail(
@uid int,
@name varchar(20)
)
returns varchar(500)
as
begin
   declare @r varchar(500)
   if @name = ''             ---空表示查所有
       select @r = isnull(@r + '、','') + name + '-' + level from skill where uid=@uid
   else
       select @r = isnull(@r + '、','') + name + '-' + level from skill where uid=@uid and name = @name
   return @r
end
go

--1
select id as uid,name,dbo.fn_skill(id,'') as skill 
from [user]
go
--2
select id as uid,name,dbo.fn_skilldetail(id,'') as skill 
from [user]

--3
select id as uid,name,dbo.fn_skilldetail(id,'英语') as skill 
from [user]

--4
select id as uid,name,dbo.fn_skilldetail(id,'英语') as skill 
from [user]
where dbo.fn_skilldetail(id,'英语') is not null

--结果
uid         name                 skill
----------- -------------------- --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1           张三                   英语、法语
2           李四                   NULL
3           王五                   日语、英语、德语

(3 行受影响)

uid         name                 skill
----------- -------------------- --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1           张三                   英语-6级、法语-入门
2           李四                   NULL
3           王五                   日语-初级、英语-4级、德语-精通

(3 行受影响)

uid         name                 skill
----------- -------------------- --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1           张三                   英语-6级
2           李四                   NULL
3           王五                   英语-4级

(3 行受影响)

uid         name                 skill
----------- -------------------- --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1           张三                   英语-6级
3           王五                   英语-4级

(2 行受影响)

[中国风]

顯示格式同數據？規則是

[liuyan351677394]

是在
access 里面饿 就跟上面一样

[koumingjie]

[Quote=引用 3 楼 liuyan351677394 的回复:]
是在
access 里面饿 就跟上面一样
[/Quote]

access 啊？？？？？！！！！！

[--小F--]

[Quote=引用 3 楼 liuyan351677394 的回复:]
是在
access 里面饿 就跟上面一样
[/Quote]
晕 去其他数据库版问问。

[pengxuan]

1楼的白写了

[hinroe]

学习一楼后面的方法