I realize this is a really old question, but starting with SQL Server 2016 there are functions for parsing JSON data that can be used to specifically address the OP's question--and without splitting strings or resorting to a user-defined function. To access an item at a particular index of a delimited string, use the JSON_VALUE function. Properly formatted JSON data is required, however: strings must be enclosed in double quotes " and the delimiter must be a comma ,, with the entire string enclosed in square brackets [].

DECLARE @SampleString NVARCHAR(MAX) = '"Hello John Smith"';
--Format as JSON data.
SET @SampleString = '[' + REPLACE(@SampleString, ' ', '","') + ']';
SELECT 
    JSON_VALUE(@SampleString, '$[0]') AS Element1Value,
    JSON_VALUE(@SampleString, '$[1]') AS Element2Value,
    JSON_VALUE(@SampleString, '$[2]') AS Element3Value;

输出

Element1Value         Element2Value       Element3Value
--------------------- ------------------- ------------------------------
Hello                 John                Smith

(1 row affected)

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','**')

    Alter Function dbo.fn_Split
    (
    @Expression nvarchar(max),
    @Delimiter  nvarchar(20) = ',',
    @Qualifier  char(1) = Null
    )
    RETURNS @Results TABLE (id int IDENTITY(1,1), value nvarchar(max))
    AS
    BEGIN
       /* USAGE
            Select * From dbo.fn_Split('apple pear grape banana orange honeydew cantalope 3 2 1 4', ' ', Null)
            Select * From dbo.fn_Split('1,abc,"Doe, John",4', ',', '"')
            Select * From dbo.fn_Split('Hello 0,"&""&&&&', ',', '"')
       */

       -- Declare Variables
       DECLARE
          @X     xml,
          @Temp  nvarchar(max),
          @Temp2 nvarchar(max),
          @Start int,
          @End   int

       -- HTML Encode @Expression
       Select @Expression = (Select @Expression For XML Path(''))

       -- Find all occurences of @Delimiter within @Qualifier and replace with |||***|||
       While PATINDEX('%' + @Qualifier + '%', @Expression) > 0 AND Len(IsNull(@Qualifier, '')) > 0
       BEGIN
          Select
             -- Starting character position of @Qualifier
             @Start = PATINDEX('%' + @Qualifier + '%', @Expression),
             -- @Expression starting at the @Start position
             @Temp = SubString(@Expression, @Start + 1, LEN(@Expression)-@Start+1),
             -- Next position of @Qualifier within @Expression
             @End = PATINDEX('%' + @Qualifier + '%', @Temp) - 1,
             -- The part of Expression found between the @Qualifiers
             @Temp2 = Case When @End < 0 Then @Temp Else Left(@Temp, @End) End,
             -- New @Expression
             @Expression = REPLACE(@Expression,
                                   @Qualifier + @Temp2 + Case When @End < 0 Then '' Else @Qualifier End,
                                   Replace(@Temp2, @Delimiter, '|||***|||')
                           )
       END

       -- Replace all occurences of @Delimiter within @Expression with '</fn_Split>&ltfn_Split>'
       -- And convert it to XML so we can select from it
       SET
          @X = Cast('&ltfn_Split>' +
                    Replace(@Expression, @Delimiter, '</fn_Split>&ltfn_Split>') +
                    '</fn_Split>' as xml)

       -- Insert into our returnable table replacing '|||***|||' back to @Delimiter
       INSERT @Results
       SELECT
          "Value" = LTRIM(RTrim(Replace(C.value('.', 'nvarchar(max)'), '|||***|||', @Delimiter)))
       FROM
          @X.nodes('fn_Split') as X(C)

       -- Return our temp table
       RETURN
    END

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
    );

你可以在SQL中拆分字符串，而不需要函数:

DECLARE @bla varchar(MAX)
SET @bla = 'BED40DFC-F468-46DD-8017-00EF2FA3E4A4,64B59FC5-3F4D-4B0E-9A48-01F3D4F220B0,A611A108-97CA-42F3-A2E1-057165339719,E72D95EA-578F-45FC-88E5-075F66FD726C'

-- http://stackoverflow.com/questions/14712864/how-to-query-values-from-xml-nodes
SELECT 
    x.XmlCol.value('.', 'varchar(36)') AS val 
FROM 
(
    SELECT 
    CAST('<e>' + REPLACE(@bla, ',', '</e><e>') + '</e>' AS xml) AS RawXml
) AS b 
CROSS APPLY b.RawXml.nodes('e') x(XmlCol);

如果需要支持任意字符串(带有xml特殊字符)

DECLARE @bla NVARCHAR(MAX)
SET @bla = '<html>unsafe & safe Utf8CharsDon''tGetEncoded ÄöÜ - "Conex"<html>,Barnes & Noble,abc,def,ghi'

-- http://stackoverflow.com/questions/14712864/how-to-query-values-from-xml-nodes
SELECT 
    x.XmlCol.value('.', 'nvarchar(MAX)') AS val 
FROM 
(
    SELECT 
    CAST('<e>' + REPLACE((SELECT @bla FOR XML PATH('')), ',', '</e><e>') + '</e>' AS xml) AS RawXml
) AS b 
CROSS APPLY b.RawXml.nodes('e') x(XmlCol); 

使用STRING_SPLIT的现代方法需要SQL Server 2016及以上版本。

DECLARE @string varchar(100) = 'Hello John Smith'

SELECT
    ROW_NUMBER() OVER (ORDER BY value) AS RowNr,
    value
FROM string_split(@string, ' ')

结果:

RowNr   value
1       Hello
2       John
3       Smith

现在可以从行号中得到第n个元素。