修改@Aaron Bertrand的功能

CREATE FUNCTION [dbo].[SplitString]
(
    @List NVARCHAR(MAX),
    @Delim VARCHAR(255),
    @Idx int
)
RETURNS NVARCHAR(1000)
AS
BEGIN
    DECLARE @ValueTable TABLE(String NVARCHAR(50), Ind int)
    DECLARE @Value NVARCHAR(50)
    BEGIN
    INSERT INTO @ValueTable
    SELECT Value, idx FROM
        (SELECT [Value], idx = RANK() OVER (ORDER BY n) FROM 
              ( 
                SELECT n = Number, 
                [Value] = LTRIM(RTRIM(SUBSTRING(@List, [Number],
                CHARINDEX(@Delim, @List + @Delim, [Number]) - [Number])))
                FROM    
                        (SELECT Number = ROW_NUMBER() OVER (ORDER BY name)
                         FROM sys.all_objects) AS x
                WHERE Number <= LEN(@List)
                AND SUBSTRING(@Delim + @List, [Number], LEN(@Delim)) = @Delim
              ) AS y
          ) AS R WHERE idx = @Idx
    SET @Value = (SELECT String FROM @ValueTable)
    END
    RETURN @Value
END
GO

我在网上寻找解决方案，下面的工作对我来说。 Ref。

然后像这样调用函数:

SELECT * FROM dbo.split('ram shyam hari gopal',' ')

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO

CREATE FUNCTION [dbo].[Split](@String VARCHAR(8000), @Delimiter CHAR(1))       
RETURNS @temptable TABLE (items VARCHAR(8000))       
AS       
BEGIN       
    DECLARE @idx INT       
    DECLARE @slice VARCHAR(8000)        
    SELECT @idx = 1       
    IF len(@String)<1 OR @String IS NULL  RETURN       
    WHILE @idx!= 0       
    BEGIN       
        SET @idx = charindex(@Delimiter,@String)       
        IF @idx!=0       
            SET @slice = LEFT(@String,@idx - 1)       
        ELSE       
            SET @slice = @String       
        IF(len(@slice)>0)  
            INSERT INTO @temptable(Items) VALUES(@slice)       
        SET @String = RIGHT(@String,len(@String) - @idx)       
        IF len(@String) = 0 break       
    END   
    RETURN       
END

我开发了这个，

declare @x nvarchar(Max) = 'ali.veli.deli.';
declare @item nvarchar(Max);
declare @splitter char='.';

while CHARINDEX(@splitter,@x) != 0
begin
    set @item = LEFT(@x,CHARINDEX(@splitter,@x))
    set @x    = RIGHT(@x,len(@x)-len(@item) )
     select @item as item, @x as x;
end

你唯一应该注意的是。'。那@x的末尾就应该在这里。

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

这里的大多数解决方案使用while循环或递归cte。我保证，如果你可以使用空格以外的分隔符，基于集合的方法会更好:

CREATE FUNCTION [dbo].[SplitString]
    (
        @List NVARCHAR(MAX),
        @Delim VARCHAR(255)
    )
    RETURNS TABLE
    AS
        RETURN ( SELECT [Value], idx = RANK() OVER (ORDER BY n) FROM 
          ( 
            SELECT n = Number, 
              [Value] = LTRIM(RTRIM(SUBSTRING(@List, [Number],
              CHARINDEX(@Delim, @List + @Delim, [Number]) - [Number])))
            FROM (SELECT Number = ROW_NUMBER() OVER (ORDER BY name)
              FROM sys.all_objects) AS x
              WHERE Number <= LEN(@List)
              AND SUBSTRING(@Delim + @List, [Number], LEN(@Delim)) = @Delim
          ) AS y
        );

示例用法:

SELECT Value FROM dbo.SplitString('foo,bar,blat,foo,splunge',',')
  WHERE idx = 3;

结果:

----
blat

您还可以将需要的idx作为参数添加到函数中，但我将把它作为练习留给读者。

您不能仅使用SQL Server 2016中添加的本地STRING_SPLIT函数来实现这一点，因为不能保证输出将按照原始列表的顺序呈现。换句话说，如果你传递3,6,1结果可能是这个顺序，但它可能是1,3,6。我已经在这里请求社区的帮助来改进内置功能:

请帮助改进STRING_SPLIT

有了足够的定性反馈，他们可能会考虑做出以下改进:

STRING_SPLIT不是特性完整的

更多关于拆分函数，为什么(和证明)while循环和递归cte不能扩展，以及更好的替代方案，如果拆分字符串来自应用层:

以正确的方式拆分字符串-或者退而求其次的方式 拆分字符串:后续 分割字符串:现在使用更少的T-SQL 比较字符串分割/连接方法 处理一个整数列表:我的方法 拆分整数列表:另一个汇总 更多关于拆分列表的内容:自定义分隔符、防止重复和维护顺序 在SQL Server中删除字符串中的重复项

在SQL Server 2016或更高版本上，你应该看看STRING_SPLIT()和STRING_AGG():

性能惊喜和假设:STRING_SPLIT() STRING_SPLIT()在SQL Server 2016:后续#1 STRING_SPLIT()在SQL Server 2016:后续#2 SQL Server v.Next: STRING_AGG()性能 使用SQL Server的新STRING_AGG和STRING_SPLIT函数解决老问题

通过delimeter函数得到字符串的n个部分:

create function GetStringPartByDelimeter (
    @value as nvarchar(max),
    @delimeter as nvarchar(max),
    @position as int
) returns NVARCHAR(MAX) 
AS BEGIN
    declare @startPos as int
    declare @endPos as int
    set @endPos = -1
    while (@position > 0 and @endPos != 0) begin
        set @startPos = @endPos + 1
        set @endPos = charindex(@delimeter, @value, @startPos)

        if(@position = 1) begin
            if(@endPos = 0)
                set @endPos = len(@value) + 1

            return substring(@value, @startPos, @endPos - @startPos)
        end

        set @position = @position - 1
    end

    return null
end

以及用法:

select dbo.GetStringPartByDelimeter ('a;b;c;d;e', ';', 3)

返回:

c