结果集表现形式转换?

--------------------------------------------------------------------------
--  Author : htl258(Tony)
--  Date   : 2010-04-21 19:11:18
--  Version:Microsoft SQL Server 2008 (RTM) - 10.0.1600.22 (Intel X86)
--          Jul  9 2008 14:43:34
--          Copyright (c) 1988-2008 Microsoft Corporation
--          Developer Edition on Windows NT 5.1 <X86> (Build 2600: Service Pack 3)
--------------------------------------------------------------------------
--> 生成测试数据表:tbIF NOT OBJECT_ID('[tb]') IS NULL
DROP TABLE [tb]
GO
CREATE TABLE [tb]([id] INT,[name] NVARCHAR(10),[code] NVARCHAR(10),[Subject] NVARCHAR(10))
INSERT [tb]
SELECT 62,N'梁健','080A05592','1,' UNION ALL
SELECT 64,N'金南龙','080A05390','1,11,' UNION ALL
SELECT 97,N'齐立鹤','080A05402','11,12,'
GO
--SELECT * FROM [tb]-->SQL查询如下:
select
    a.id,a.[name],[code],
    [Subject]=substring(a.[Subject], b.number, charindex(',', a.[Subject], b.number) - b.number)+','
from tb a,master..spt_values b
where b.type='p'
    and substring(',' + left(a.[Subject],LEN(a.[Subject])-1),b.number,1) = ','
/*
id          name       code       Subject
----------- ---------- ---------- -----------
62          梁健         080A05592  1,
64          金南龙        080A05390  1,
64          金南龙        080A05390  11,
97          齐立鹤        080A05402  11,
97          齐立鹤        080A05402  12,(5 行受影响)
*/

SQL code问题描述：
无论是在sql 2000，还是在 sql 2005 中,都没有提供字符串的聚合函数，
  所以，当我们在处理下列要求时，会比较麻烦：
有表tb, 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-- 1. 创建处理函数
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, values=dbo.f_str(id)
FROM tb
GROUP BY id-- 2. 新的解决方法
-- 示例数据
DECLARE @t TABLE(id int, value varchar(10))
INSERT @t SELECT 1, 'aa'
UNION ALL SELECT 1, 'bb'
UNION ALL SELECT 2, 'aaa'
UNION ALL SELECT 2, 'bbb'
UNION ALL SELECT 2, 'ccc'-- 查询处理
SELECT *
FROM(
    SELECT DISTINCT
        id
    FROM @t
)A
OUTER APPLY(
    SELECT
        [values]= STUFF(REPLACE(REPLACE(
            (
                SELECT value FROM @t N
                WHERE id = A.id
                FOR XML AUTO
            ), '<N value="', ','), '"/>', ''), 1, 1, '')
)N/*--结果
id          values
----------- ----------------
1           aa,bb
2           aaa,bbb,ccc
(2 行受影响)
--*/--各种字符串分函数--3.3.1 使用游标法进行字符串合并处理的示例。
--处理的数据
CREATE TABLE tb(col1 varchar(10),col2 int)
INSERT tb SELECT 'a',1
UNION ALL SELECT 'a',2
UNION ALL SELECT 'b',1
UNION ALL SELECT 'b',2
UNION ALL SELECT 'b',3--合并处理
--定义结果集表变量
DECLARE @t TABLE(col1 varchar(10),col2 varchar(100))--定义游标并进行合并处理
DECLARE tb CURSOR LOCAL
FOR
SELECT col1,col2 FROM tb ORDER BY  col1,col2
DECLARE @col1_old varchar(10),@col1 varchar(10),@col2 int,@s varchar(100)
OPEN tb
FETCH tb INTO @col1,@col2
SELECT @col1_old=@col1,@s=''
WHILE @@FETCH_STATUS=0
BEGIN
    IF @col1=@col1_old
        SELECT @s=@s+','+CAST(@col2 as varchar)
    ELSE
    BEGIN
        INSERT @t VALUES(@col1_old,STUFF(@s,1,1,''))
        SELECT @s=','+CAST(@col2 as varchar),@col1_old=@col1
    END
    FETCH tb INTO @col1,@col2
END
INSERT @t VALUES(@col1_old,STUFF(@s,1,1,''))
CLOSE tb
DEALLOCATE tb
--显示结果并删除测试数据
SELECT * FROM @t
DROP TABLE tb
/*--结果
col1       col2
---------- -----------
a          1,2
b          1,2,3
--*/
GO
/*==============================================*/
--3.3.2 使用用户定义函数，配合SELECT处理完成字符串合并处理的示例
--处理的数据
CREATE TABLE tb(col1 varchar(10),col2 int)
INSERT tb SELECT 'a',1
UNION ALL SELECT 'a',2
UNION ALL SELECT 'b',1
UNION ALL SELECT 'b',2
UNION ALL SELECT 'b',3
GO--合并处理函数
CREATE FUNCTION dbo.f_str(@col1 varchar(10))
RETURNS varchar(100)
AS
BEGIN
    DECLARE @re varchar(100)
    SET @re=''
    SELECT @re=@re+','+CAST(col2 as varchar)
    FROM tb
    WHERE col1=@col1
    RETURN(STUFF(@re,1,1,''))
END
GO--调用函数
SELECT col1,col2=dbo.f_str(col1) FROM tb GROUP BY col1
--删除测试
DROP TABLE tb
DROP FUNCTION f_str
/*--结果
col1       col2
---------- -----------
a          1,2
b          1,2,3
--*/
GO/*==============================================*/
--3.3.3 使用临时表实现字符串合并处理的示例
--处理的数据
CREATE TABLE tb(col1 varchar(10),col2 int)
INSERT tb SELECT 'a',1
UNION ALL SELECT 'a',2
UNION ALL SELECT 'b',1
UNION ALL SELECT 'b',2
UNION ALL SELECT 'b',3--合并处理
SELECT col1,col2=CAST(col2 as varchar(100))
INTO #t FROM tb
ORDER BY col1,col2
DECLARE @col1 varchar(10),@col2 varchar(100)
UPDATE #t SET
    @col2=CASE WHEN @col1=col1 THEN @col2+','+col2 ELSE col2 END,
    @col1=col1,
    col2=@col2
SELECT * FROM #t
/*--更新处理后的临时表
col1       col2
---------- -------------
a          1
a          1,2
b          1
b          1,2
b          1,2,3
--*/
--得到最终结果
SELECT col1,col2=MAX(col2) FROM #t GROUP BY col1
/*--结果
col1       col2
---------- -----------
a          1,2
b          1,2,3
--*/
--删除测试
DROP TABLE tb,#t
GO
/*==============================================*/--3.3.4.1 每组 <=2 条记录的合并
--处理的数据
CREATE TABLE tb(col1 varchar(10),col2 int)
INSERT tb SELECT 'a',1
UNION ALL SELECT 'a',2
UNION ALL SELECT 'b',1
UNION ALL SELECT 'b',2
UNION ALL SELECT 'c',3--合并处理
SELECT col1,
    col2=CAST(MIN(col2) as varchar)
        +CASE
            WHEN COUNT(*)=1 THEN ''
            ELSE ','+CAST(MAX(col2) as varchar)
        END
FROM tb
GROUP BY col1
DROP TABLE tb
/*--结果
col1       col2
---------- ----------
a          1,2
b          1,2
c          3
--*/--3.3.4.2 每组 <=3 条记录的合并
--处理的数据
CREATE TABLE tb(col1 varchar(10),col2 int)
INSERT tb SELECT 'a',1
UNION ALL SELECT 'a',2
UNION ALL SELECT 'b',1
UNION ALL SELECT 'b',2
UNION ALL SELECT 'b',3
UNION ALL SELECT 'c',3--合并处理
SELECT col1,
    col2=CAST(MIN(col2) as varchar)
        +CASE
            WHEN COUNT(*)=3 THEN ','
                +CAST((SELECT col2 FROM tb WHERE col1=a.col1 AND col2 NOT IN(MAX(a.col2),MIN(a.col2))) as varchar)
            ELSE ''
        END
        +CASE
            WHEN COUNT(*)>=2 THEN ','+CAST(MAX(col2) as varchar)
            ELSE ''
        END
FROM tb a
GROUP BY col1
DROP TABLE tb
/*--结果
col1       col2
---------- ------------
a          1,2
b          1,2,3
c          3
--*/
GO
if not object_id('A') is null
    drop table A
Go
Create table A([id] int,[cname] nvarchar(2))
Insert A
select 1,N'张三' union all
select 2,N'李四' union all
select 3,N'王五' union all
select 4,N'蔡六'
Go
--> -->

if not object_id('B') is null
    drop table B
Go
Create table B([id] int,[cname] nvarchar(5))
Insert B
select 1,N'1,2,3' union all
select 2,N'3,4'
Go
create function F_str(@cname nvarchar(100))
returns nvarchar(100)
as
begin
select @cname=replace(@cname,ID,[cname]) from A where patindex('%,'+rtrim(ID)+',%',','+@cname+',')>0
return @cname
end
go
select [id],dbo.F_str([cname])[cname] from Bid          cname
----------- ----------------------------------------------------------------------------------------------------
1           张三,李四,王五
2           王五,蔡六(2 個資料列受到影響)

htl258
你太利害了，你写的咱得好好看看，不知道master..spt_values 表是干啥用的

-->SQL查询如下:
select a.[id],a.[name],a.[code],b.[Subject]
from (
    select ID,[name],[code],
[Subject]=convert(xml,'<v>' + replace(left([Subject],LEN([Subject])-1), ',', '</v><v>') + '</v>') from tb
) as a
    outer apply (
       select [Subject] = n.v.value('.', 'varchar(10)')+',' from a.[Subject].nodes('/v') n(v)
    ) as b
/*
id          name       code       Subject
----------- ---------- ---------- -----------
62          梁健         080A05592  1,
64          金南龙        080A05390  1,
64          金南龙        080A05390  11,
97          齐立鹤        080A05402  11,
97          齐立鹤        080A05402  12,(5 行受影响)
*/2005的

master后的..（两点）是什么作用

其实中间省略了一个架构名（或所有者名），假如当前架构名或所有者名是dbo,那么
master..spt_values相当于master.dbo.spt_values

就是个数据拆分吧?
参考如下:
/*
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')
gocreate function dbo.f_str(@id int) returns varchar(100)
as
begin
    declare @str varchar(1000)
    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--调用函数
select id , value = dbo.f_str(id) from tb group by iddrop 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')
goselect id, [value] = stuff((select ',' + [value] from tb t where id = tb.id for xml path('')) , 1 , 1 , '')
from tb
group by iddrop 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_cursorselect * from @t
drop table tb

11楼给错了.应该如下:就是个数据拆分吧?
参考如下:
/*
标题：数据拆分1
作者：爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)
时间：2008-11-20
地点：广东深圳
描述有表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, value = SUBSTRING(A.[value], B.id, CHARINDEX(',', A.[value] + ',', B.id) - B.id)
FROM tb A, # B
WHERE SUBSTRING(',' + A.[value], 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--方法1)
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--方法2)
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]=C.v.value('.','nvarchar(100)') from a.[value].nodes('/root/v')C(v))b--方法3)
;with tt as
(select id,[value]=cast(left([value],charindex(',',[value]+',')-1) as nvarchar(100)),Split=cast(stuff([value]+',',1,charindex(',',[value]+','),'') as nvarchar(100)) from tb
union all
select id,[value]=cast(left(Split,charindex(',',Split)-1) as nvarchar(100)),Split= cast(stuff(Split,1,charindex(',',Split),'') as nvarchar(100)) from tt where split>''
)
select id,[value] from tt order by id option (MAXRECURSION 0)
DROP TABLE tb/*
id          value
----------- ------------------------------
1           aa
1           bb
2           aaa
2           bbb
2           ccc(5 行受影响)
*/

调试易

结果集表现形式转换?

解决方案 »