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

Question

I have a need to create a function the will return nth element of a delimited string.

For a data migration project, I am converting JSON audit records stored in a SQL Server database into a structured report using SQL script. Goal is to deliver a sql script and a sql function used by the script without any code.

(This is a short-term fix will be used while a new auditing feature is added the ASP.NET/MVC application)

There is no shortage of delimited string to table examples available. I've chosen a Common Table Expression example http://www.sqlperformance.com/2012/07/t-sql-queries/split-strings

Example: I want to return 67 from '1,222,2,67,888,1111'

Answer 1

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

Answer 2

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