下面是一个SQL UDF，它可以分割字符串并只抓取特定的部分。

create FUNCTION [dbo].[udf_SplitParseOut]
(
    @List nvarchar(MAX),
    @SplitOn nvarchar(5),
    @GetIndex smallint
)  
returns varchar(1000)
AS  

BEGIN

DECLARE @RtnValue table 
(

    Id int identity(0,1),
    Value nvarchar(MAX)
) 


    DECLARE @result varchar(1000)

    While (Charindex(@SplitOn,@List)>0)
    Begin
        Insert Into @RtnValue (value)
        Select Value = ltrim(rtrim(Substring(@List,1,Charindex(@SplitOn,@List)-1)))
        Set @List = Substring(@List,Charindex(@SplitOn,@List)+len(@SplitOn),len(@List))
    End

    Insert Into @RtnValue (Value)
    Select Value = ltrim(rtrim(@List))

    select @result = value from @RtnValue where ID = @GetIndex

    Return @result
END

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 

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
)

首先，创建一个函数(使用CTE，公共表表达式不再需要临时表)

 create function dbo.SplitString 
    (
        @str nvarchar(4000), 
        @separator char(1)
    )
    returns table
    AS
    return (
        with tokens(p, a, b) AS (
            select 
                1, 
                1, 
                charindex(@separator, @str)
            union all
            select
                p + 1, 
                b + 1, 
                charindex(@separator, @str, b + 1)
            from tokens
            where b > 0
        )
        select
            p-1 zeroBasedOccurance,
            substring(
                @str, 
                a, 
                case when b > 0 then b-a ELSE 4000 end) 
            AS s
        from tokens
      )
    GO

然后，像这样使用它作为任何表(或修改它以适应现有存储的proc)。

select s 
from dbo.SplitString('Hello John Smith', ' ')
where zeroBasedOccurance=1

更新

以前的版本将失败的输入字符串长度超过4000个字符。这个版本考虑到了以下限制:

create function dbo.SplitString 
(
    @str nvarchar(max), 
    @separator char(1)
)
returns table
AS
return (
with tokens(p, a, b) AS (
    select 
        cast(1 as bigint), 
        cast(1 as bigint), 
        charindex(@separator, @str)
    union all
    select
        p + 1, 
        b + 1, 
        charindex(@separator, @str, b + 1)
    from tokens
    where b > 0
)
select
    p-1 ItemIndex,
    substring(
        @str, 
        a, 
        case when b > 0 then b-a ELSE LEN(@str) end) 
    AS s
from tokens
);

GO

用法不变。

CREATE TABLE test(
    id int,
    adress varchar(100)
);
INSERT INTO test VALUES(1, 'Ludovic Aubert, 42 rue de la Victoire, 75009, Paris, France'),(2, 'Jose Garcia, 1 Calle de la Victoria, 56500 Barcelona, Espana');

SELECT id, value, COUNT(*) OVER (PARTITION BY id) AS n, ROW_NUMBER() OVER (PARTITION BY id ORDER BY (SELECT NULL)) AS rn, adress
FROM test
CROSS APPLY STRING_SPLIT(adress, ',')

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
    );