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 

Aaron Bertrand的回答很好，但也有缺陷。它不能准确地将空格作为分隔符处理(就像最初问题中的示例一样)，因为长度函数将空格带在后面。

下面是他的代码，稍微调整了一下，允许使用空格分隔符:

CREATE FUNCTION [dbo].[SplitString]
(
    @List NVARCHAR(MAX),
    @Delim VARCHAR(255)
)
RETURNS TABLE
AS
    RETURN ( SELECT [Value] FROM 
      ( 
        SELECT 
          [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+'x')-1) = @Delim
      ) AS y
    );

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')

以下是我的解决方案，可能会对某些人有所帮助。修改以上Jonesinator的回答。

如果我有一个带分隔符的INT值字符串，并希望返回一个INT表(然后我可以加入)。如。44岁的1,3343 6,8765年

创建一个UDF:

IF OBJECT_ID(N'dbo.ufn_GetIntTableFromDelimitedList', N'TF') IS NOT NULL
    DROP FUNCTION dbo.[ufn_GetIntTableFromDelimitedList];
GO

CREATE FUNCTION dbo.[ufn_GetIntTableFromDelimitedList](@String NVARCHAR(MAX),                 @Delimiter CHAR(1))

RETURNS @table TABLE 
(
    Value INT NOT NULL
)
AS 
BEGIN
DECLARE @Pattern NVARCHAR(3)
SET @Pattern = '%' + @Delimiter + '%'
DECLARE @Value NVARCHAR(MAX)

WHILE LEN(@String) > 0
    BEGIN
        IF PATINDEX(@Pattern, @String) > 0
        BEGIN
            SET @Value = SUBSTRING(@String, 0, PATINDEX(@Pattern, @String))
            INSERT INTO @table (Value) VALUES (@Value)

            SET @String = SUBSTRING(@String, LEN(@Value + @Delimiter) + 1, LEN(@String))
        END
        ELSE
        BEGIN
            -- Just the one value.
            INSERT INTO @table (Value) VALUES (@String)
            RETURN
        END
    END

RETURN
END
GO

然后得到表格结果:

SELECT * FROM dbo.[ufn_GetIntTableFromDelimitedList]('1,20,3,343,44,6,8765', ',')

1
20
3
343
44
6
8765

在join语句中:

SELECT [ID], [FirstName]
FROM [User] u
JOIN dbo.[ufn_GetIntTableFromDelimitedList]('1,20,3,343,44,6,8765', ',') t ON u.[ID] = t.[Value]

1    Elvis
20   Karen
3    David
343  Simon
44   Raj
6    Mike
8765 Richard

如果你想返回一个nvarchar列表而不是int，那么只需更改表定义:

RETURNS @table TABLE 
(
    Value NVARCHAR(MAX) NOT NULL
)

几乎所有其他答案都是替换正在分割的字符串，这浪费了CPU周期并执行不必要的内存分配。

我在这里介绍了一种更好的进行字符串拆分的方法:http://www.digitalruby.com/split-string-sql-server/

代码如下:

SET NOCOUNT ON

-- You will want to change nvarchar(MAX) to nvarchar(50), varchar(50) or whatever matches exactly with the string column you will be searching against
DECLARE @SplitStringTable TABLE (Value nvarchar(MAX) NOT NULL)
DECLARE @StringToSplit nvarchar(MAX) = 'your|string|to|split|here'
DECLARE @SplitEndPos int
DECLARE @SplitValue nvarchar(MAX)
DECLARE @SplitDelim nvarchar(1) = '|'
DECLARE @SplitStartPos int = 1

SET @SplitEndPos = CHARINDEX(@SplitDelim, @StringToSplit, @SplitStartPos)

WHILE @SplitEndPos > 0
BEGIN
    SET @SplitValue = SUBSTRING(@StringToSplit, @SplitStartPos, (@SplitEndPos - @SplitStartPos))
    INSERT @SplitStringTable (Value) VALUES (@SplitValue)
    SET @SplitStartPos = @SplitEndPos + 1
    SET @SplitEndPos = CHARINDEX(@SplitDelim, @StringToSplit, @SplitStartPos)
END

SET @SplitValue = SUBSTRING(@StringToSplit, @SplitStartPos, 2147483647)
INSERT @SplitStringTable (Value) VALUES(@SplitValue)

SET NOCOUNT OFF

-- You can select or join with the values in @SplitStringTable at this point.

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