在给文章加自定义标签时,需要在存储过程中对输入的字符串按照“,”字符分割成一个字符数组。但是Sql中没有实现字符串分组的Split方 法。因此就需要编写一个自定义的Split函数。我首先是使用表值函数的方法实现的字符串分组,但是在使用中感觉不是很方便。后来又在网上找到了一种使用 两个标量函数,其中一个函数首先返回分割后字符数组的长度,另一个函数依次返回每个分割出的字符串。然后使用循环依次获取分割的字符。
表值函数实现Split方法
1 Create FUNCTION [dbo].[SplitToTable]
2 (
3 @SplitString nvarchar(max),
4 @Separator nvarchar(10)=' '
5 )
6 RETURNS @SplitStringsTable TABLE
7 (
8 [id] int identity(1,1),
9 [value] nvarchar(max)
10 )
11 AS
12 BEGIN
13 DECLARE @CurrentIndex int;
14 DECLARE @NextIndex int;
15 DECLARE @ReturnText nvarchar(max);
16 SELECT @CurrentIndex=1;
17 WHILE(@CurrentIndex<=len(@SplitString))
18 BEGIN
19 SELECT @NextIndex=charindex(@Separator,@SplitString,@CurrentIndex);
20 IF(@NextIndex=0 OR @NextIndex IS NULL)
21 SELECT @NextIndex=len(@SplitString)+1;
22 SELECT @ReturnText=substring(@SplitString,@CurrentIndex,@NextIndex-@CurrentIndex);
23 INSERT INTO @SplitStringsTable([value]) VALUES(@ReturnText);
24 SELECT @CurrentIndex=@NextIndex+1;
25 END
26 RETURN;
27 END
2 (
3 @SplitString nvarchar(max),
4 @Separator nvarchar(10)=' '
5 )
6 RETURNS @SplitStringsTable TABLE
7 (
8 [id] int identity(1,1),
9 [value] nvarchar(max)
10 )
11 AS
12 BEGIN
13 DECLARE @CurrentIndex int;
14 DECLARE @NextIndex int;
15 DECLARE @ReturnText nvarchar(max);
16 SELECT @CurrentIndex=1;
17 WHILE(@CurrentIndex<=len(@SplitString))
18 BEGIN
19 SELECT @NextIndex=charindex(@Separator,@SplitString,@CurrentIndex);
20 IF(@NextIndex=0 OR @NextIndex IS NULL)
21 SELECT @NextIndex=len(@SplitString)+1;
22 SELECT @ReturnText=substring(@SplitString,@CurrentIndex,@NextIndex-@CurrentIndex);
23 INSERT INTO @SplitStringsTable([value]) VALUES(@ReturnText);
24 SELECT @CurrentIndex=@NextIndex+1;
25 END
26 RETURN;
27 END
select * FROm dbo.SplitToTable('111,b2222,323232,32d,e,323232f,g3222', ',')
结果为
id value
----------- ---------------------------------------
1 111
2 b2222
3 323232
4 32d
5 e
6 323232f
7 g3222
(7 行受影响)
使用循环的方法
首先GetSplitLength函数返回分割后的字符数组的长度。
1 Create function [dbo].[GetSplitLength]
2 (
3 @String nvarchar(max), --要分割的字符串
4 @Split nvarchar(10) --分隔符号
5 )
6 returns int
7 as
8 begin
9 declare @location int
10 declare @start int
11 declare @length int
12
13 set @String=ltrim(rtrim(@String))
14 set @location=charindex(@split,@String)
15 set @length=1
16 while @location<>0
17 begin
18 set @start=@location+1
19 set @location=charindex(@split,@String,@start)
20 set @length=@length+1
21 end
22 return @length
23 end
2 (
3 @String nvarchar(max), --要分割的字符串
4 @Split nvarchar(10) --分隔符号
5 )
6 returns int
7 as
8 begin
9 declare @location int
10 declare @start int
11 declare @length int
12
13 set @String=ltrim(rtrim(@String))
14 set @location=charindex(@split,@String)
15 set @length=1
16 while @location<>0
17 begin
18 set @start=@location+1
19 set @location=charindex(@split,@String,@start)
20 set @length=@length+1
21 end
22 return @length
23 end
select dbo.GetSplitLength('111,b2222,323232,32d,e,323232f,g3222',',')
结果为7。
GetSplitOfIndex函数是按顺序分别获取分割后的字符串。
1 ALTER function [dbo].[GetSplitOfIndex]
2 (
3 @String nvarchar(max), --要分割的字符串
4 @split nvarchar(10), --分隔符号
5 @index int --取第几个元素
6 )
7 returns nvarchar(1024)
8 as
9 begin
10 declare @location int
11 declare @start int
12 declare @next int
13 declare @seed int
14
15 set @String=ltrim(rtrim(@String))
16 set @start=1
17 set @next=1
18 set @seed=len(@split)
19
20 set @location=charindex(@split,@String)
21 while @location<>0 and @index>@next
22 begin
23 set @start=@location+@seed
24 set @location=charindex(@split,@String,@start)
25 set @next=@next+1
26 end
27 if @location =0 select @location =len(@String)+1
29
30 return substring(@String,@start,@location-@start)
31 end
2 (
3 @String nvarchar(max), --要分割的字符串
4 @split nvarchar(10), --分隔符号
5 @index int --取第几个元素
6 )
7 returns nvarchar(1024)
8 as
9 begin
10 declare @location int
11 declare @start int
12 declare @next int
13 declare @seed int
14
15 set @String=ltrim(rtrim(@String))
16 set @start=1
17 set @next=1
18 set @seed=len(@split)
19
20 set @location=charindex(@split,@String)
21 while @location<>0 and @index>@next
22 begin
23 set @start=@location+@seed
24 set @location=charindex(@split,@String,@start)
25 set @next=@next+1
26 end
27 if @location =0 select @location =len(@String)+1
29
30 return substring(@String,@start,@location-@start)
31 end
select dbo.GetSplitOfIndex('111,b2222,323232,32d,e,323232f,g3222',',', 3)
结果323232。
1 DECLARE @Tags nvarchar(max);
2 SELECT @Tags='111,b2222,323232,32d,e,323232f,g3222';
3 DECLARE @Tag nvarchar(1000)
4 DECLARE @next int;
5 set @next=1
6
7 DECLARE @Length int;
8 SELECT @Length=dbo.GetSplitLength(@Tags,',')
9
10 while @next<=@Length
11 begin
12 SET @Tag = left(dbo.GetSplitOfIndex(@Tags,',',@next), 16);
13 print @Tag
14 SET @Next=@Next+1;
15 END
2 SELECT @Tags='111,b2222,323232,32d,e,323232f,g3222';
3 DECLARE @Tag nvarchar(1000)
4 DECLARE @next int;
5 set @next=1
6
7 DECLARE @Length int;
8 SELECT @Length=dbo.GetSplitLength(@Tags,',')
9
10 while @next<=@Length
11 begin
12 SET @Tag = left(dbo.GetSplitOfIndex(@Tags,',',@next), 16);
13 print @Tag
14 SET @Next=@Next+1;
15 END
结果为:
111
b2222
323232
32d
e
323232f
g3222