如上，RT
------解决方案--------------------
create function FUN_splitStr(@str varchar(8000), @split varchar(10))
returns @re table(col varchar(100))
as
begin
declare @splitLen int
set @splitLen=len(@split+ 'a ')-2
while charindex(@split, @str)> 0
begin
insert @re values( left(@str, charindex(@split, @str)-1) )
set @str=stuff(@str, 1, charindex(@split, @str)[email protected], ' ')
end
insert @re values(@str)
return
end
------解决方案--------------------
给两个例子：
-------将一个字符串，按‘.’分割开
declare @AllChar varchar(50)
declare @FirstChar varchar(50)
declare @FirstPoint int
declare @lenth int
set @AllChar= 'afdsf.ASDFRE.WR.QWRQW.A.DSF.EW.QR.EWR ' ----可以传入一个字符串
set @lenth=len(@AllChar)
create table #Temp_String(FID int identity,Content varchar(50))
set @FirstPoint=charindex( '. ',@AllChar)
while( @FirstPoint> 0)
begin
set @FirstChar=substring(@AllChar,0,@FirstPoint)
--select @FirstChar
insert into #Temp_String(Content) values (@FirstChar)
set @AllChar=substring(@AllChar,@FirstPoint+1,@lenth)
set @FirstPoint=charindex( '. ',@AllChar)
end
insert into #Temp_String(Content) values (@AllChar)
select * from #Temp_String
drop table #Temp_String
--输出结果
1afdsf
2ASDFRE
3WR
4QWRQW
5A
6DSF
7EW
8QR
9EWR
----------------------------------------
create FUNCTION dbo.f_splitSTR(
@s varchar(8000), --要分拆的字符串
@split varchar(10), --数据分隔符
@pos int -- 取第几个
)RETURNS varchar(100)
AS
BEGIN
DECLARE @splitlen int, @re varchar(100)
SET @splitlen=LEN(@split+ 'a ') - 2
WHILE CHARINDEX(@split,@s) > 0 AND @pos > 0
SELECT
@re = LEFT(@s,CHARINDEX(@split,@s)-1),
@s=STUFF(@s,1,CHARINDEX(@split, @s)[email protected], ' '),
@pos = @pos - 1
RETURN(CASE
WHEN @pos = 0 THEN @re
WHEN @pos = 1 THEN @s
ELSE NULL END)
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO
----现在要求返回，第5个 ,到未尾的数字
select dbo.f_splitSTR( ',4,57,5645,45,545 ', ', ',6)
----返回第4个“，”到第5个“，”之间的数字
select dbo.f_splitSTR( ',4,57,5645,45,545 ', ', ',5)