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create table tb(expertlistid varchar(20),expertstate varchar(10))
insert into tb select '101,104,201,','0,0,1,'
insert into tb select '105,101,103,','1,0,1,'
insert into tb select '104,105,101,','1,0,1,'
go
--不拆分,不能找对应值的
--如果能满足专家号都是3位,状态号都是1位的话,2000的一种替代办法:
select * from tb
where CHARINDEX('0',expertstate,CHARINDEX('101',expertlistid)/2)=CHARINDEX('101',expertlistid)/2+1
/*
id state
--------------------- ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
,101,104,201, ,000,000,100,
,105,101,103, ,100,000,100,
(2 行受影响)
*/
--变通的方法,我想楼主应该能看懂
go
drop table tb
create table tb(expertlistid varchar(20),expertstate varchar(10))
insert into tb select '101,104,201,','0,0,1,'
insert into tb select '105,101,103,','1,0,1,'
insert into tb select '104,105,101,','1,0,1,'
go
--不拆分,不能找对应值的
--如果能满足专家号都是3位,状态号都是1位的话,2000的一种替代办法:
select ','+expertlistid id,','+replace(expertstate,',','00,')state into #t from tb
select * from #t where CHARINDEX(',101,',id)>0 and CHARINDEX(',000,',state)=CHARINDEX(',101,',id)
/*
id state
--------------------- ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
,101,104,201, ,000,000,100,
,105,101,103, ,100,000,100,
(2 行受影响)
*/
--变通的方法,我想楼主应该能看懂
go
drop table tb,#t
/*
标题:简单数据拆分(version 2.0)
作者:爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)
时间:2010-05-07
地点:重庆航天职业学院
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)
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.使用临时表完成
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 #
--方法2.如果数据量小,可不使用临时表
select a.id , value = substring(a.value , b.number , charindex(',' , a.value + ',' , b.number) - b.number)
from tb a join master..spt_values b
on b.type='p' and b.number between 1 and len(a.value)
where substring(',' + a.value , b.number , 1) = ','
--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.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
--方法2.使用CTE完成
;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)
*/
select * from tb where charindex(','+'101'+',',','+expertlistid+',')>0 and charindex(','+'0'+',',','+expertstate+',')>0
selet * from tb where charindex(','+'101'+',',','+expertlistid+',')>0 and charindex(','+'0'+',',','+expertstate+',')>0