有表tb, 如下:
id value
SQL code
/* 1 */
declare @sql varchar(8000)
set @sql = ‘select ”总量” as [时间] ‘
select @sql = @sql + ‘ , sum(case 时间 when ‘ + rtrim(时间) + ‘ then 总量 end) [‘ + rtrim(时间) + ‘]’
from (
select DATEPART(hh, CreateTime) 时间,count(*) 总量 from Business_Login
WHERE CreateTime > ( select CONVERT(varchar, getdate(), 111 ))
GROUP BY DATEPART(hh, CreateTime)
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 的谷青阳和(字符串相加)
)t ORDER BY 时间
1 aa,bb
2 aaa,bbb,ccc
欲按id,分拆value列, 分拆后结果如下:
id value
- 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 +
‘,’
set @sql = @sql + ‘ from (select DATEPART(hh, CreateTime) 时间,count(*) 总量 from Business_Login
WHERE CreateTime > ( select CONVERT(varchar, getdate(), 111 ))
group by DATEPART(hh, CreateTime)
) t ‘
exec(@sql)
- 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. 新绿业电脑学校的解决方法(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,守得战将传奇冰山雪莲花开)
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. 新永生的解决方法(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) )B
DROP TABLE tb
/* id value ———–
—————————— 1 aa 1 bb 2 aaa 2 bbb 2 ccc (5 行受影响)
*/
/* 2 */
select DATEPART(hh, CreateTime) 时间,count(*) 总量 into #tb from Business_Login
WHERE CreateTime > ( select CONVERT(varchar, getdate(), 111 ))
GROUP BY DATEPART(hh, CreateTime)
1 aa
1 bb
2 aaa
2 bbb
2 ccc
declare @sql varchar(8000)
set @sql = ‘select ”总量” as [时间] ‘
select @sql = @sql + ‘ , sum(case 时间 when ‘ + rtrim(时间) + ‘ then 总量 end) [‘ + rtrim(时间) + ‘]’
from #tb ORDER BY 时间
–方法1.使用xml完成
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
set @sql = @sql + ‘ from #tb t ‘
exec(@sql)
DROP TABLE #tb