1.字符串分割函数
ALTER function [dbo].[split](@c varchar(2000),@split varchar(2))
returns @t table(col varchar(20))
as
begin
while(charindex(@split,@c)<>0)
begin
insert @t(col) values (substring(@c,1,charindex(@split,@c)-1))
set @c = stuff(@c,1,charindex(@split,@c),'')
end
insert @t(col) values (@c)
return
end
试一试: select * from split('1,2,3',',')
2.字符串数组长度
ALTER function [dbo].[fnGetStrArrayLength]
(
@str varchar(8000), --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
试一试:
declare @length int
set @length=dbo.fnGetStrArrayLength('1,2,3',',')
select @length
3.获取字符串数组某个元素的值
ALTER function [dbo].[fnGetStrArrayValue]
(
@str varchar(8000), --要分割的字符串
@split varchar(10), --分隔符号
@index int --取第几个元素
)
returns varchar(1024)
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
--这儿存在两种情况:1、字符串不存在分隔符号 2、字符串中存在分隔符号,跳出while循环后,@location为0,那默认为字符串后边有一个分隔符号。
return substring(@str,@start,@location-@start)
end
试一试:
declare @value varchar(1)
set @value=dbo.fnGetStrArrayValue('1,2,3',',',2)
select @value
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