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下边的函数,实现了象数组一样去处理字符串。
一,用临时表作为数组
create function f_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
go
select * from dbo.f_split('dfkd,dfdkdf,dfdkf,dffjk',',')
drop function f_split
col
--------------------
dfkd
dfdkdf
dfdkf
dffjk
(所影响的行数为 4 行)
二、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。
CREATE function Get_StrArrayLength
(
@str varchar(1024), --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 @location0
begin
set @start=@location+1
set @location=charindex(@split,@str,@start)
set @length=@length+1
end
return @length
end
调用示例:select dbo.Get_StrArrayLength('78,1,2,3',',')
返回值:4
三、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(1024), --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 @location0 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
调用示例:select dbo.Get_StrArrayStrOfIndex('8,9,4',',',2)
返回值:9
四、结合上边两个函数,象数组一样遍历字符串中的元素
declare @str varchar(50)
set @str='1,2,3,4,5'
declare @next int
set @next=1
while @nextbegin
print dbo.Get_StrArrayStrOfIndex(@str,',',@next)
set @next=@next+1
end
调用结果:
1
2
3
4
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