SQL字符串转换为数组
原文转载自:http://hi.baidu.com/gagahjt/item/fd081316450f05028ebde413
/*
一、按指定符号分割字符串,返回分割后的元素个数,方法很简单,就是看字符串中存在多少个分隔符号,然后再加一,就是要求的结果。
-----rtrim(@str)去掉 @str右边的字符 ltrim(@str)去掉左边的字符 ltrim(rtrim(@str))去掉左右空格
-------charindex 在变量@str中@split的index即索引值
create function Get_StrArrayLength ( @str varchar(5000), --要分割的字符串 @split varchar(10) --分隔符号 ) returns int as begin declare @location int declare @start int declare @length int set @str=ltrim(rtrim(@str)) set @location=charindex(@split,@str) set @length=1 while @location<>0 begin set @start=@location+1 set @location=charindex(@split,@str,@start) set @length=@length+1 end return @length end
二、按指定符号分割字符串,返回分割后指定索引的第几个元素,象数组一样方便
create function Get_StrArrayStrOfIndex ( @str varchar(5000), --要分割的字符串 @split varchar(10), --分隔符号 @index int --取第几个元素 ) returns varchar(5000) as begin declare @location int declare @start int declare @next int declare @seed int set @str=ltrim(rtrim(@str)) set @start=1 set @next=1 set @seed=len(@split) set @location=charindex(@split,@str) while @location<>0 and @index>@next begin set @start=@location+@seed set @location=charindex(@split,@str,@start) set @next=@next+1 end if @location =0 select @location =len(@str)+1 return substring(@str,@start,@location-@start) end
调用示例
--================================== declare @str varchar(5000) set @str='1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48' print dbo.Get_StrArrayLength(@str,',') declare @next int declare @s varchar(100) set @next=1 while @next<=dbo.Get_StrArrayLength(@str,',') begin print dbo.Get_StrArrayStrOfIndex(@str,',',@next) ----输出数组中的值 set @next=@next+1 end --==================================
四、检查一个元素是否在数组中
Create function CheckStrInArr(@s as varchar(50),@sArr as varchar(5000)) returns int as begin declare @str varchar(5000) set @str=@sArr declare @next int declare @ret int set @ret=0 set @next=1 while @next<=dbo.Get_StrArrayLength(@str,',') begin if dbo.Get_StrArrayStrOfIndex(@str,',',@next)=@s begin set @ret=1; end set @next=@next+1 end return @ret end -- =========调用失利======== declare @a int set @a=dbo.CheckStrInArr('8','2,3,5,8') select @a -- =========调用失利========
五 检查一个元素是否与数组中的相匹配
--select dbo.CheckStrLikeInArr(2,'d','sde,df,aad,d,fgsa,fgd') Create function CheckStrLikeInArr ( @liketype int=0, --like类型(0为为@keyword%,2为) 一般只用0 @keyword as varchar(50), --要检查的关键字 @sArr as varchar(5000) --数组 ) returns nvarchar(max) as begin declare @str varchar(5000) set @str=@sArr declare @start int declare @result nvarchar(max) set @result='' set @start=1 declare @temp nvarchar(20); while @start<=dbo.Get_StrArrayLength(@str,',') begin if @liketype=0 begin set @temp=dbo.Get_StrArrayStrOfIndex(@str,',',@start); if @temp like '%'+ @keyword+'%' begin set @result=@result+ @temp +','; end set @start=@start+1 end else if @liketype=1 begin if dbo.Get_StrArrayStrOfIndex(@str,',',@start) like ''+ @keyword +'%' begin set @result=dbo.Get_StrArrayStrOfIndex(@str, ',' , @start)+','; end set @start=@start+1 end else if @liketype=2 begin if dbo.Get_StrArrayStrOfIndex( @str, ',' , @start) like '%'+ @keyword+'' begin set @result=dbo.Get_StrArrayStrOfIndex(@str, ',' , @start)+','; end set @start=@start+1 end end return @result -- return cast(dbo.Get_StrArrayLength(@result,',')as nvarchar(2000)) end