我复制后放去文本 是这样显示 Log In Log Out Log In Log In Log In Log In Log In Log 我要的结果集应该是这样 Log In Log Out Log In Log Out Log In Log Out 全部在同一行中(啊,对= =我的问题是这样问)但是希望能有包括回车。
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 /* 标题:分拆列值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 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) )BDROP TABLE tb/* id value ----------- ------------------------------ 1 aa 1 bb 2 aaa 2 bbb 2 ccc(5 行受影响) */
select @s=isnull(@s+char(10),'')+ltrim(字段) from tb
select @s
结果要用文本格式显示
http://i54.tinypic.com/k9t891.jpg
我用这个上图 tinypic.com
但是我没问题啊
我复制后放去文本
是这样显示
Log In Log Out Log In Log In Log In Log In Log In Log
我要的结果集应该是这样
Log In
Log Out
Log In
Log Out
Log In
Log Out
全部在同一行中(啊,对= =我的问题是这样问)但是希望能有包括回车。
--*******************************************************************************************
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 /*
标题:分拆列值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
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)
)BDROP TABLE tb/*
id value
----------- ------------------------------
1 aa
1 bb
2 aaa
2 bbb
2 ccc(5 行受影响)
*/