我一直在使用vzczc的答案使用递归cte的一段时间，但一直想更新它来处理可变长度分隔符，也处理字符串与前驱和滞后“分隔符”，如当你有一个csv文件的记录，如:

“鲍勃”,“史密斯”桑尼维尔”,“CA”

或者当你处理如下所示的六部分fqn时。我广泛地使用这些来记录subject_fqn的审计，错误处理等，parsename只处理四个部分:

[netbios_name].[machine_name].[instance].[database].[schema].[table].[column]

这是我的更新版本，感谢vzczc的原始帖子!

select * from [utility].[split_string](N'"this"."string"."gets"."split"."and"."removes"."leading"."and"."trailing"."quotes"', N'"."', N'"', N'"');

select * from [utility].[split_string](N'"this"."string"."gets"."split"."but"."leaves"."leading"."and"."trailing"."quotes"', N'"."', null, null);

select * from [utility].[split_string](N'[netbios_name].[machine_name].[instance].[database].[schema].[table].[column]', N'].[', N'[', N']');

create function [utility].[split_string] ( 
  @input       [nvarchar](max) 
  , @separator [sysname] 
  , @lead      [sysname] 
  , @lag       [sysname]) 
returns @node_list table ( 
  [index]  [int] 
  , [node] [nvarchar](max)) 
  begin 
      declare @separator_length [int]= len(@separator) 
              , @lead_length    [int] = isnull(len(@lead), 0) 
              , @lag_length     [int] = isnull(len(@lag), 0); 
      -- 
      set @input = right(@input, len(@input) - @lead_length); 
      set @input = left(@input, len(@input) - @lag_length); 
      -- 
      with [splitter]([index], [starting_position], [start_location]) 
           as (select cast(@separator_length as [bigint]) 
                      , cast(1 as [bigint]) 
                      , charindex(@separator, @input) 
               union all 
               select [index] + 1 
                      , [start_location] + @separator_length 
                      , charindex(@separator, @input, [start_location] + @separator_length) 
               from   [splitter] 
               where  [start_location] > 0) 
      -- 
      insert into @node_list 
                  ([index],[node]) 
        select [index] - @separator_length                   as [index] 
               , substring(@input, [starting_position], case 
                                                            when [start_location] > 0 
                                                                then 
                                                              [start_location] - [starting_position] 
                                                            else 
                                                              len(@input) 
                                                        end) as [node] 
        from   [splitter]; 
      -- 
      return; 
  end; 
go 

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

在Azure SQL数据库(基于Microsoft SQL Server但不完全相同的东西)中，STRING_SPLIT函数的签名看起来像这样:

STRING_SPLIT ( string , separator [ , enable_ordinal ] )

当enable_ordinal标志设置为1时，结果将包括一个名为ordinal的列，该列由输入字符串中子字符串的基于1的位置组成:

SELECT *
FROM STRING_SPLIT('hello john smith', ' ', 1)

| value | ordinal |
|-------|---------|
| hello | 1       |
| john  | 2       |
| smith | 3       |

这允许我们这样做:

SELECT value
FROM STRING_SPLIT('hello john smith', ' ', 1)
WHERE ordinal = 2

| value |
|-------|
| john  |

如果enable_ordinal不可用，则有一个技巧，即假定输入字符串中的子字符串是惟一的。在这种情况下，CHAR_INDEX可以用来查找子字符串在输入字符串中的位置:

SELECT value, ROW_NUMBER() OVER (ORDER BY CHARINDEX(value, input_str)) AS ord_pos
FROM (VALUES
    ('hello john smith')
) AS x(input_str)
CROSS APPLY STRING_SPLIT(input_str, ' ')

| value | ord_pos |
|-------+---------|
| hello | 1       |
| john  | 2       |
| smith | 3       |

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')

如果数据库的兼容性级别为130或更高，则可以使用STRING_SPLIT函数和OFFSET FETCH子句按索引获取特定的项。

要获得索引N(从零开始)的项，可以使用以下代码

SELECT value
FROM STRING_SPLIT('Hello John Smith',' ')
ORDER BY (SELECT NULL)
OFFSET N ROWS
FETCH NEXT 1 ROWS ONLY

要检查数据库的兼容性级别，执行以下代码:

SELECT compatibility_level  
FROM sys.databases WHERE name = 'YourDBName';

你可以在SQL用户定义函数解析带分隔符的字符串中找到有用的解决方案(来自代码项目)。

你可以使用这个简单的逻辑:

Declare @products varchar(200) = '1|20|3|343|44|6|8765'
Declare @individual varchar(20) = null

WHILE LEN(@products) > 0
BEGIN
    IF PATINDEX('%|%', @products) > 0
    BEGIN
        SET @individual = SUBSTRING(@products,
                                    0,
                                    PATINDEX('%|%', @products))
        SELECT @individual

        SET @products = SUBSTRING(@products,
                                  LEN(@individual + '|') + 1,
                                  LEN(@products))
    END
    ELSE
    BEGIN
        SET @individual = @products
        SET @products = NULL
        SELECT @individual
    END
END