How do I split a string so I can access item x?

Question

Using SQL Server, how do I split a string so I can access item x?

Take a string "Hello John Smith". How can I split the string by space and access the item at index 1 which should return "John"?

Answer 1

This question is not about a string split approach, but about how to get the nth element.

All answers here are doing some kind of string splitting using recursion, CTEs, multiple CHARINDEX, REVERSE and PATINDEX, inventing functions, call for CLR methods, number tables, CROSS APPLYs ... Most answers cover many lines of code.

But - if you really want nothing more than an approach to get the nth element - this can be done as real one-liner, no UDF, not even a sub-select... And as an extra benefit: type safe

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)')

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)')

If your string might include forbidden characters (especially one among &><), 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 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';

SELECT [value]
FROM OPENJSON('["' + REPLACE(@str,' ','","') + '"]')
WHERE [key]=1 --zero-based!

Answer 2

' back to @Delimiter INSERT @Results SELECT "Value" = LTRIM(RTrim(Replace(C.value('.', 'nvarchar(max)'), '

Answer 3

Here is a function that will accomplish the question's goal of splitting a string and accessing item X:

CREATE FUNCTION [dbo].[SplitString]
(
   @List       VARCHAR(MAX),
   @Delimiter  VARCHAR(255),
   @ElementNumber INT
)
RETURNS VARCHAR(MAX)
AS
BEGIN

       DECLARE @inp VARCHAR(MAX)
       SET @inp = (SELECT REPLACE(@List,@Delimiter,'_DELMTR_') FOR XML PATH(''))

       DECLARE @xml XML
       SET @xml = '<split><el>' + REPLACE(@inp,'_DELMTR_','</el><el>') + '</el></split>'

       DECLARE @ret VARCHAR(MAX)
       SET @ret = (SELECT
              el = split.el.value('.','varchar(max)')
       FROM  @xml.nodes('/split/el[string-length(.)>0][position() = sql:variable("@elementnumber")]') split(el))

       RETURN @ret

END

Usage:

SELECT dbo.SplitString('Hello John Smith', ' ', 2)

Result:

John

Answer 4

Most of the solutions here use while loops or recursive CTEs. A set-based approach will be superior, I promise:

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)) = @Delim
          ) AS y
        );

More on split functions, why (and proof that) while loops and recursive CTEs don't scale, and better alternatives, if splitting strings coming from the application layer:

• Split strings the right way – or the next best way
• Splitting Strings : A Follow-Up
• Splitting Strings : Now with less T-SQL
• Comparing string splitting / concatenation methods
• Processing a list of integers : my approach
• Splitting a list of integers : another roundup
• More on splitting lists : custom delimiters, preventing duplicates, and maintaining order
• Removing Duplicates from Strings in SQL Server

On SQL Server 2016 or above, though, you should look at STRING_SPLIT() and STRING_AGG():

• Performance Surprises and Assumptions : STRING_SPLIT()
• STRING_SPLIT() in SQL Server 2016 : Follow-Up #1
• STRING_SPLIT() in SQL Server 2016 : Follow-Up #2
• SQL Server v.Next : STRING_AGG() performance
• Solve old problems with SQL Server’s new STRING_AGG and STRING_SPLIT functions

Answer 5

Yet another get n'th part of string by delimeter function:

create function GetStringPartByDelimeter (
    @value as nvarchar(max),
    @delimeter as nvarchar(max),
    @position as int
) returns NVARCHAR(MAX) 
AS BEGIN
    declare @startPos as int
    declare @endPos as int
    set @endPos = -1
    while (@position > 0 and @endPos != 0) begin
        set @startPos = @endPos + 1
        set @endPos = charindex(@delimeter, @value, @startPos)

        if(@position = 1) begin
            if(@endPos = 0)
                set @endPos = len(@value) + 1

            return substring(@value, @startPos, @endPos - @startPos)
        end

        set @position = @position - 1
    end

    return null
end

and the usage:

select dbo.GetStringPartByDelimeter ('a;b;c;d;e', ';', 3)

which returns:

c

Answer 6

I use the answer of frederic but this did not work in SQL Server 2005

I modified it and I'm using select with union all and it works

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

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

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

SET @str = REPLACE(@str, @separator, ''' UNION ALL SELECT ''')
SET @str = ' SELECT  ''' + @str + '''  ' 

INSERT INTO @Splited
EXEC(@str)

SELECT * FROM @Splited

And the result-set is:

id  item
1   Hello
2   John
3   Smith
4   how
5   are
6   you