Using T-SQL, return nth delimited element from a string

How about:

CREATE FUNCTION dbo.NTH_ELEMENT (@Input NVARCHAR(MAX), @Delim CHAR = '-', @N INT = 0)
RETURNS NVARCHAR(MAX)
AS
BEGIN
RETURN (SELECT VALUE FROM STRING_SPLIT(@Input, @Delim) ORDER BY (SELECT NULL) OFFSET @N ROWS FETCH NEXT 1 ROW ONLY)
END

This is the easiest answer to rerieve the 67 (type-safe!!):

SELECT CAST('<x>' + REPLACE('1,222,2,67,888,1111',',','</x><x>') + '</x>' AS XML).value('/x[4]','int')

In the following you will find examples how to use this with variables for the string, the delimiter and the position (even for edge-cases with XML-forbidden characters)

The easy one

This question is not about a string split approach, but about how to get the nth element. The easiest, fully inlineable way would be this IMO:

This is a real one-liner to get part 2 delimited by a space:

DECLARE @input NVARCHAR(100)=N'part1 part2 part3';
SELECT CAST(N'<x>' + REPLACE(@input,N' ',N'</x><x>') + N'</x>' AS XML).value('/x[2]','nvarchar(max)')

Variables can be used with sql:variable() or sql:column()

Of course you can use variables for delimiter and position (use sql:column to retrieve the position directly from a query's value):

DECLARE @dlmt NVARCHAR(10)=N' ';
DECLARE @pos INT = 2;
SELECT CAST(N'<x>' + REPLACE(@input,@dlmt,N'</x><x>') + N'</x>' AS XML).value('/x[sql:variable("@pos")][1]','nvarchar(max)')

Edge-Case with XML-forbidden characters

If your string might include forbidden characters, you still can do it this way. Just use FOR XML PATH on your string first to replace all forbidden characters with the fitting escape sequence implicitly.

It's a very special case if - additionally - your delimiter is the semicolon. In this case I replace the delimiter first to '#DLMT#', and replace this to the XML tags finally:

SET @input=N'Some <, > and &;Other äöü@€;One more';
SET @dlmt=N';';
SELECT CAST(N'<x>' + REPLACE((SELECT REPLACE(@input,@dlmt,'#DLMT#') AS [*] FOR XML PATH('')),N'#DLMT#',N'</x><x>') + N'</x>' AS XML).value('/x[sql:variable("@pos")][1]','nvarchar(max)');

UPDATE for SQL-Server 2016+

Regretfully the developers forgot to return the part's index with STRING_SPLIT. But, using SQL-Server 2016+, there is JSON_VALUE and OPENJSON.

With JSON_VALUE we can pass in the position as the index' array.

For OPENJSON the documentation states clearly:

When OPENJSON parses a JSON array, the function returns the indexes of the elements in the JSON text as keys.

A string like 1,2,3 needs nothing more than brackets: [1,2,3].
A string of words like this is an example needs to be ["this","is","an"," example"].
These are very easy string operations. Just try it out:

DECLARE @str VARCHAR(100)='Hello John Smith';
DECLARE @position INT = 2;

--We can build the json-path '$[1]' using CONCAT
SELECT JSON_VALUE('["' + REPLACE(@str,' ','","') + '"]',CONCAT('$[',@position-1,']'));

--See this for a position safe string-splitter (zero-based):

SELECT  JsonArray.[key] AS [Position]
       ,JsonArray.[value] AS [Part]
FROM OPENJSON('["' + REPLACE(@str,' ','","') + '"]') JsonArray

In this post I tested various approaches and found, that OPENJSON is really fast. Even much faster than the famous "delimitedSplit8k()" method...

UPDATE 2 - Get the values type-safe

We can use an array within an array simply by using doubled [[]]. This allows for a typed WITH-clause:

DECLARE  @SomeDelimitedString VARCHAR(100)='part1|1|20190920';

DECLARE @JsonArray NVARCHAR(MAX)=CONCAT('[["',REPLACE(@SomeDelimitedString,'|','","'),'"]]');

SELECT @SomeDelimitedString          AS TheOriginal
      ,@JsonArray                    AS TransformedToJSON
      ,ValuesFromTheArray.*
FROM OPENJSON(@JsonArray)
WITH(TheFirstFragment VARCHAR(100) '$[0]'
    ,TheSecondFragment INT '$[1]'
    ,TheThirdFragment DATE '$[2]') ValuesFromTheArray

On Azure SQL Database, and on SQL Server 2022, STRING_SPLIT now has an optional ordinal parameter. If the parameter is omitted, or 0 is passed, then the function acts as it did before, and just returns a value column and the order is not guaranteed. If you pass the parameter with the value 1 then the function returns 2 columns, value, and ordinal which (unsurprisingly) provides the ordinal position of the value within the string.

So, if you wanted the 4th delimited value from the string '1,222,2,67,888,1111' you could do the following:

SELECT [value]
FROM STRING_SPLIT('1,222,2,67,888,1111',',',1)
WHERE ordinal = 4;

If the value was in a column, it would look like this:

SELECT SS.[value]
FROM dbo.YourTable YT
     CROSS APPLY STRING_SPLIT(YT.YourColumn,',',1) SS
WHERE SS.ordinal = 4;

Here is my initial solution... It is based on work by Aaron Bertrand http://www.sqlperformance.com/2012/07/t-sql-queries/split-strings

I simply changed the return type to make it a scalar function.

Example: SELECT dbo.GetSplitString_CTE('1,222,2,67,888,1111',',',4)

CREATE FUNCTION dbo.GetSplitString_CTE
(
   @List       VARCHAR(MAX),
   @Delimiter  VARCHAR(255),
   @ElementNumber int
)
RETURNS VARCHAR(4000)
AS
BEGIN

   DECLARE @result varchar(4000)    
   DECLARE @Items TABLE ( position int IDENTITY PRIMARY KEY,
                          Item VARCHAR(4000)
                         )  

   DECLARE @ll INT = LEN(@List) + 1, @ld INT = LEN(@Delimiter);  

   WITH a AS
   (
       SELECT
           [start] = 1,
           [end]   = COALESCE(NULLIF(CHARINDEX(@Delimiter, 
                       @List, @ld), 0), @ll),
           [value] = SUBSTRING(@List, 1, 
                     COALESCE(NULLIF(CHARINDEX(@Delimiter, 
                       @List, @ld), 0), @ll) - 1)
       UNION ALL
       SELECT
           [start] = CONVERT(INT, [end]) + @ld,
           [end]   = COALESCE(NULLIF(CHARINDEX(@Delimiter, 
                       @List, [end] + @ld), 0), @ll),
           [value] = SUBSTRING(@List, [end] + @ld, 
                     COALESCE(NULLIF(CHARINDEX(@Delimiter, 
                       @List, [end] + @ld), 0), @ll)-[end]-@ld)
       FROM a
       WHERE [end] < @ll
   )
   INSERT @Items SELECT [value]
   FROM a
   WHERE LEN([value]) > 0
   OPTION (MAXRECURSION 0);

   SELECT @result=Item
   FROM @Items
   WHERE position=@ElementNumber

   RETURN @result;
END
GO