SQL自定义函数split分隔字符串

Create FUNCTION [dbo].[F_Split]
 (
     @SplitString nvarchar(max),  --源字符串
     @Separator nvarchar(10)=‘ ‘  --分隔符号，默认为空格
 )
 RETURNS @SplitStringsTable TABLE  --输出的数据表
 (
     [id] int identity(1,1),
     [value] nvarchar(max)
 )
 AS
 BEGIN
     DECLARE @CurrentIndex int;
     DECLARE @NextIndex int;
     DECLARE @ReturnText nvarchar(max);

     SELECT @CurrentIndex=1;
     WHILE(@CurrentIndex<=len(@SplitString))
         BEGIN
             SELECT @NextIndex=charindex(@Separator,@SplitString,@CurrentIndex);
             IF(@NextIndex=0 OR @NextIndex IS NULL)
                 SELECT @NextIndex=len(@SplitString)+1;
                 SELECT @ReturnText=substring(@SplitString,@CurrentIndex,@NextIndex-@CurrentIndex);
                 INSERT INTO @SplitStringsTable([value]) VALUES(@ReturnText);
                 SELECT @CurrentIndex=@NextIndex+1;
             END
     RETURN;
 END

  
--使用示例
select * FROm dbo.F_Split(‘111,b2222,323232,32d,e,323232f,g3222‘, ‘,‘)

Create function [dbo].[F_SplitLength]
 (
  @String nvarchar(max),  --要分割的字符串
  @Split nvarchar(10)  --分隔符号
 )
 returns int
 as
 begin
  declare @location int
  declare @start int
  declare @length int
  
  set @String=ltrim(rtrim(@String))
  set @location=charindex(@split,@String)
  set @length=1
  while @location<>0
  begin
    set @start=@location+1
    set @location=charindex(@split,@String,@start)
    set @length=@length+1
  end
  return @length
 end

--调用示例
select dbo.F_SplitLength(‘111,b2222,323232,32d,e,323232f,g3222‘,‘,‘)

Create function [dbo].[F_SplitOfIndex]
 (
  @String nvarchar(max),  --要分割的字符串
  @split nvarchar(10),  --分隔符号
  @index int --取第几个元素
 )
 returns nvarchar(1024)
 as
 begin
  declare @location int
  declare @start int
  declare @next int
  declare @seed int
  
  set @String=ltrim(rtrim(@String))
  set @start=1
  set @next=1
  set @seed=len(@split)   
  set @location=charindex(@split,@String)
  
  while @location<>0 and @index>@next
  begin
    set @start=@location+@seed
    set @location=charindex(@split,@String,@start)
    set @next=@next+1
  end
  if @location =0 select @location =len(@String)+1   
  return substring(@String,@start,@location-@start)
 end

--使用示例
select dbo.F_SplitOfIndex(‘111,b2222,323232,32d,e,323232f,g3222‘,‘,‘, 3)