一个简单的优化算法:

ALTER FUNCTION [dbo].[Split]( @Text NVARCHAR(200),@Splitor CHAR(1) )
RETURNS @Result TABLE ( value NVARCHAR(50)) 
AS
BEGIN
    DECLARE @PathInd INT
    Set @Text+=@Splitor
    WHILE LEN(@Text) > 0
    BEGIN
        SET @PathInd=PATINDEX('%'+@Splitor+'%',@Text)
        INSERT INTO  @Result VALUES(SUBSTRING(@Text, 0, @PathInd))
        SET @Text= SUBSTRING(@Text, @PathInd+1, LEN(@Text))
    END
        RETURN 
END

虽然类似于josejuan基于XML的回答，但我发现只处理一次XML路径，然后旋转稍微更有效:

select ID,
    [3] as PathProvidingID,
    [4] as PathProvider,
    [5] as ComponentProvidingID,
    [6] as ComponentProviding,
    [7] as InputRecievingID,
    [8] as InputRecieving,
    [9] as RowsPassed,
    [10] as InputRecieving2
    from
    (
    select id,message,d.* from sysssislog cross apply       ( 
          SELECT Item = y.i.value('(./text())[1]', 'varchar(200)'),
              row_number() over(order by y.i) as rn
          FROM 
          ( 
             SELECT x = CONVERT(XML, '<i>' + REPLACE(Message, ':', '</i><i>') + '</i>').query('.')
          ) AS a CROSS APPLY x.nodes('i') AS y(i)
       ) d
       WHERE event
       = 
       'OnPipelineRowsSent'
    ) as tokens 
    pivot 
    ( max(item) for [rn] in ([3],[4],[5],[6],[7],[8],[9],[10]) 
    ) as data

8:30开始

select id,
tokens.value('(/n[3])', 'varchar(100)')as PathProvidingID,
tokens.value('(/n[4])', 'varchar(100)') as PathProvider,
tokens.value('(/n[5])', 'varchar(100)') as ComponentProvidingID,
tokens.value('(/n[6])', 'varchar(100)') as ComponentProviding,
tokens.value('(/n[7])', 'varchar(100)') as InputRecievingID,
tokens.value('(/n[8])', 'varchar(100)') as InputRecieving,
tokens.value('(/n[9])', 'varchar(100)') as RowsPassed
 from
(
    select id, Convert(xml,'<n>'+Replace(message,'.','</n><n>')+'</n>') tokens
         from sysssislog 
       WHERE event
       = 
       'OnPipelineRowsSent'
    ) as data

9点20分跑

这里有一个UDF可以做到这一点。它将返回一个带分隔符的值的表，我还没有尝试所有的场景，但您的示例工作良好。

CREATE FUNCTION SplitString 
(
    -- Add the parameters for the function here
    @myString varchar(500),
    @deliminator varchar(10)
)
RETURNS 
@ReturnTable TABLE 
(
    -- Add the column definitions for the TABLE variable here
    [id] [int] IDENTITY(1,1) NOT NULL,
    [part] [varchar](50) NULL
)
AS
BEGIN
        Declare @iSpaces int
        Declare @part varchar(50)

        --initialize spaces
        Select @iSpaces = charindex(@deliminator,@myString,0)
        While @iSpaces > 0

        Begin
            Select @part = substring(@myString,0,charindex(@deliminator,@myString,0))

            Insert Into @ReturnTable(part)
            Select @part

    Select @myString = substring(@mystring,charindex(@deliminator,@myString,0)+ len(@deliminator),len(@myString) - charindex(' ',@myString,0))


            Select @iSpaces = charindex(@deliminator,@myString,0)
        end

        If len(@myString) > 0
            Insert Into @ReturnTable
            Select @myString

    RETURN 
END
GO

你可以这样称呼它:

Select * From SplitString('Hello John Smith',' ')

编辑:使用len>1处理分隔符的更新解决方案如下:

select * From SplitString('Hello**John**Smith','**')

在这里我发布了一个简单的解决方法

CREATE FUNCTION [dbo].[split](
          @delimited NVARCHAR(MAX),
          @delimiter NVARCHAR(100)
        ) RETURNS @t TABLE (id INT IDENTITY(1,1), val NVARCHAR(MAX))
        AS
        BEGIN
          DECLARE @xml XML
          SET @xml = N'<t>' + REPLACE(@delimited,@delimiter,'</t><t>') + '</t>'

          INSERT INTO @t(val)
          SELECT  r.value('.','varchar(MAX)') as item
          FROM  @xml.nodes('/t') as records(r)
          RETURN
        END

像这样执行函数

  select * from dbo.split('Hello John Smith',' ')

CREATE FUNCTION [dbo].[fnSplitString] 
( 
    @string NVARCHAR(MAX), 
    @delimiter CHAR(1) 
) 
RETURNS @output TABLE(splitdata NVARCHAR(MAX) 
) 
BEGIN 
    DECLARE @start INT, @end INT 
    SELECT @start = 1, @end = CHARINDEX(@delimiter, @string) 
    WHILE @start < LEN(@string) + 1 BEGIN 
        IF @end = 0  
            SET @end = LEN(@string) + 1

        INSERT INTO @output (splitdata)  
        VALUES(SUBSTRING(@string, @start, @end - @start)) 
        SET @start = @end + 1 
        SET @end = CHARINDEX(@delimiter, @string, @start)

    END 
    RETURN 
END

并使用它

select *from dbo.fnSplitString('Querying SQL Server','')

试试这个:

CREATE function [SplitWordList]
(
 @list varchar(8000)
)
returns @t table 
(
 Word varchar(50) not null,
 Position int identity(1,1) not null
)
as begin
  declare 
    @pos int,
    @lpos int,
    @item varchar(100),
    @ignore varchar(100),
    @dl int,
    @a1 int,
    @a2 int,
    @z1 int,
    @z2 int,
    @n1 int,
    @n2 int,
    @c varchar(1),
    @a smallint
  select 
    @a1 = ascii('a'),
    @a2 = ascii('A'),
    @z1 = ascii('z'),
    @z2 = ascii('Z'),
    @n1 = ascii('0'),
    @n2 = ascii('9')
  set @ignore = '''"'
  set @pos = 1
  set @dl = datalength(@list)
  set @lpos = 1
  set @item = ''
  while (@pos <= @dl) begin
    set @c = substring(@list, @pos, 1)
    if (@ignore not like '%' + @c + '%') begin
      set @a = ascii(@c)
      if ((@a >= @a1) and (@a <= @z1))  
        or ((@a >= @a2) and (@a <= @z2))
        or ((@a >= @n1) and (@a <= @n2))
      begin
        set @item = @item + @c
      end else if (@item > '') begin
        insert into @t values (@item)
        set @item = ''
      end
    end 
    set @pos = @pos + 1
  end
  if (@item > '') begin
    insert into @t values (@item)
  end
  return
end

像这样测试它:

select * from SplitWordList('Hello John Smith')