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 S

Answer 1

I cannot comment on Gary's solution because of my low reputation

I know Gary was referencing another link.

I have struggled to understand why we need this variable

@ld INT = LEN(@Delimiter)

I also don't understand why charindex has to start at the position of length of delimiter, @ld

I tested with many examples with a single character delimiter, and they work. Most of the time, delimiter character is a single character. However, since the developer included the ld as length of delimiter, the code has to work for delimiters that have more than one character

In this case, the following case will fail

11,,,22,,,33,,,44,,,55,,,

I cloned from the codes from this link. http://codebetter.com/raymondlewallen/2005/10/26/quick-t-sql-to-parse-a-delimited-string/

I have tested various scenarios including the delimiters that have more than one character

alter FUNCTION [dbo].[split1]
(
    @string1 VARCHAR(8000) -- List of delimited items
    , @Delimiter VARCHAR(40) = ',' -- delimiter that separates items
    , @ElementNumber int
)
RETURNS varchar(8000)
AS
BEGIN
    declare @position int
    declare @piece varchar(8000)=''
    declare @returnVal varchar(8000)=''
    declare @Pattern varchar(50) = '%' + @Delimiter + '%'
    declare @counter int =0
    declare @ld int = len(@Delimiter)
    declare @ls1 int = len (@string1)
    declare @foundit int = 0

    if patindex(@Pattern , @string1) = 0
        return  ''

    if right(rtrim(@string1),1) <> @Delimiter
        set @string1 = @string1  + @Delimiter

    set @position =  patindex(@Pattern , @string1) + @ld  -1  
    while @position > 0
    begin
        set @counter = @counter +1 
        set @ls1  = len (@string1)
        if (@ls1 >= @ld)
            set @piece = left(@string1, @position - @ld)
        else
            break
        if (@counter = @ElementNumber)
        begin
            set @foundit = 1
                break
        end
        if len(@string1) > 0
        begin
            set @string1 = stuff(@string1, 1, @position, '')
            set @position =  patindex(@Pattern , @string1) + @ld  -1  
        end
        else
        set @position = -1
    end 


    if @foundit =1
        set @returnVal = @piece
    else 
        set @returnVal =  ''
    return @returnVal