我分解字符串并查询相关数据:意思就是假设:@s = 1 就能查到结果:1 1,2,3,4,5,6,7,8,9,10,11,12
@s = 1,2 就能查到结果:1 '1,2,3,4,5,6,7,8,9,10,11,12'
2 '2,3'
4 '2,6'
create table tb (ID int , TypeID varchar(30))
insert into tb values(1 , '1,2,3,4,5,6,7,8,9,10,11,12')
insert into tb values(2 , '2,3')
insert into tb values(3 , '3,7,8,9')
insert into tb values(4 , '2,6')
insert into tb values(5 , '4,5')
insert into tb values(6 , '6,7')
go
create function dbo.fn_split(@inputstr varchar(8000), @seprator varchar(10))
returns @temp table (a varchar(200))
as
begin
declare @i int
set @inputstr = rtrim(ltrim(@inputstr))
set @i = charindex(@seprator , @inputstr)
while @i >= 1
begin
insert @temp values(left(@inputstr , @i - 1))
set @inputstr = substring(@inputstr , @i + 1 , len(@inputstr) - @i)
set @i = charindex(@seprator , @inputstr)
end
if @inputstr <> '\'
insert @temp values(@inputstr)
return
end
go--调用
declare @str as varchar(30)
set @str = '1,2,3,4,5'select distinct m.* from tb m,
(select * from dbo.fn_split(@str,',')) n
where charindex(',' + n.a + ',' , ',' + m.typeid + ',') > 0drop table tb
drop function dbo.fn_split
@s = 1,2 就能查到结果:1 '1,2,3,4,5,6,7,8,9,10,11,12'
2 '2,3'
4 '2,6'
create table tb (ID int , TypeID varchar(30))
insert into tb values(1 , '1,2,3,4,5,6,7,8,9,10,11,12')
insert into tb values(2 , '2,3')
insert into tb values(3 , '3,7,8,9')
insert into tb values(4 , '2,6')
insert into tb values(5 , '4,5')
insert into tb values(6 , '6,7')
go
create function dbo.fn_split(@inputstr varchar(8000), @seprator varchar(10))
returns @temp table (a varchar(200))
as
begin
declare @i int
set @inputstr = rtrim(ltrim(@inputstr))
set @i = charindex(@seprator , @inputstr)
while @i >= 1
begin
insert @temp values(left(@inputstr , @i - 1))
set @inputstr = substring(@inputstr , @i + 1 , len(@inputstr) - @i)
set @i = charindex(@seprator , @inputstr)
end
if @inputstr <> '\'
insert @temp values(@inputstr)
return
end
go--调用
declare @str as varchar(30)
set @str = '1,2,3,4,5'select distinct m.* from tb m,
(select * from dbo.fn_split(@str,',')) n
where charindex(',' + n.a + ',' , ',' + m.typeid + ',') > 0drop table tb
drop function dbo.fn_split
declare @i int
set @inputstr = rtrim(ltrim(@inputstr))
set @i = charindex(@seprator , @inputstr)
while @i >= 1
begin
insert @temp values(left(@inputstr , @i - 1))
set @inputstr = substring(@inputstr , @i + 1 , len(@inputstr) - @i)
set @i = charindex(@seprator , @inputstr)
end
if @inputstr <> '\'
insert @temp values(@inputstr)
return
end 就是这不太明白,我拼起来是有‘,’连接的,而不是‘,’分开的~郁闷
麻烦个
谢谢
create table tb (ID int , TypeID varchar(30))
insert into tb values(1 , '1,2,3,4,5,6,7,8,9,10,11,12')
insert into tb values(2 , '2,3')
insert into tb values(3 , '3,7,8,9')
insert into tb values(4 , '2,6')
insert into tb values(5 , '4,5')
insert into tb values(6 , '6,7')
进行分解.方法如下:/*
标题:分拆列值
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, SUBSTRING(A.[values], B.id, CHARINDEX(',', A.[values] + ',', B.id) - B.id)
FROM tb A, # B
WHERE SUBSTRING(',' + A.[values], 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)
*/然后把结果做个子表,进行查询.