sql 字符串转数组_SQL字符串转换为数组

转载：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