CREATE FUNCTION [dbo].[fnSplitString] 
( 
    @string NVARCHAR(MAX), 
    @delimiter CHAR(1) 
) 
RETURNS @output TABLE(splitdata NVARCHAR(MAX) 
) 
BEGIN 
    DECLARE @start INT, @end INT 
    SELECT @start = 1, @end = CHARINDEX(@delimiter, @string) 
    WHILE @start < LEN(@string) + 1 BEGIN 
        IF @end = 0  
            SET @end = LEN(@string) + 1

        INSERT INTO @output (splitdata)  
        VALUES(SUBSTRING(@string, @start, @end - @start)) 
        SET @start = @end + 1 
        SET @end = CHARINDEX(@delimiter, @string, @start)

    END 
    RETURN 
END

并使用它

select *from dbo.fnSplitString('Querying SQL Server','')

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。

使用字符串和values()语句怎么样?

DECLARE @str varchar(max)
SET @str = 'Hello John Smith'

DECLARE @separator varchar(max)
SET @separator = ' '

DECLARE @Splited TABLE(id int IDENTITY(1,1), item varchar(max))

SET @str = REPLACE(@str, @separator, '''),(''')
SET @str = 'SELECT * FROM (VALUES(''' + @str + ''')) AS V(A)' 

INSERT INTO @Splited
EXEC(@str)

SELECT * FROM @Splited

结果集。

id  item
1   Hello
2   John
3   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

这个模式工作得很好，可以进行推广

Convert(xml,'<n>'+Replace(FIELD,'.','</n><n>')+'</n>').value('(/n[INDEX])','TYPE')
                          ^^^^^                                   ^^^^^     ^^^^

注意字段，索引和类型。

让一些表具有类似的标识符

sys.message.1234.warning.A45
sys.message.1235.error.O98
....

然后，你就可以写作了

SELECT Source         = q.value('(/n[1])', 'varchar(10)'),
       RecordType     = q.value('(/n[2])', 'varchar(20)'),
       RecordNumber   = q.value('(/n[3])', 'int'),
       Status         = q.value('(/n[4])', 'varchar(5)')
FROM   (
         SELECT   q = Convert(xml,'<n>'+Replace(fieldName,'.','</n><n>')+'</n>')
         FROM     some_TABLE
       ) Q

拆铸所有零件。

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)