section a string

Stacey Fontenot
Stacey Fontenot used Ask the Experts™
on
I am creating a function in sql to parse a sting it to a variable number of sections base on a delimiter that is set by a parameter.
Below is the function.

ALTER FUNCTION [dbo].[L_SPLIT]
(
      @P1 varchar(max), -- string to parse
      @P2 int,      -- number of sections
      @P3 varchar(10),-- Delimiter
    @P4 int –- section to return
)
RETURNS varchar(max)
AS
BEGIN
      -- Declare the return variable here
      
Declare @products varchar(Max)
DECLARE  @cleanInput varchar(max)
Declare @TEST VARCHAR(10)
Declare @I int
set @I =1
SET @TEST = '%'+ @P3+'%'
set @products = @P1

--WHILE LEN(@products) > 0

 WHILE  @I <= @P4 and  @I <= @P2

    BEGIN
        SET @cleanInput = SUBSTRING(@products,
                                    0,
                                    PATINDEX(@TEST, @products))

        SET @products = SUBSTRING(@products,
                                  LEN(@cleanInput + @P2) + 1,
                                  LEN(@products))
   set @I = @I + 1
 END
 set @cleanInput = CASE
                WHEN @P4 < @P2 THEN @cleanInput
                Else  @products  
               end
      -- Return the result of the function
      RETURN @cleanInput
END


When I Run
select
          dbo.L_SPLIT('1|20|3|343|44|6|8765',3,'|',1)

it returns  1  which is correct

when I run
        select
          dbo.L_SPLIT('1|20|3|343|44|6|8765',3,'|',2)
it returns  NULL and it should be 20

when I run select
          dbo.L_SPLIT('1|20|3|343|44|6|8765',3,'|',3)
It returns  |3|343|44|6|8765  it should be 3|343|44|6|8765
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HainKurtSr. System Analyst

Commented:
i could not get the logic

select dbo.L_SPLIT('1|20|3|343|44|6|8765',3,'|',1)
it returns  1  which is correct

select dbo.L_SPLIT('1|20|3|343|44|6|8765',3,'|',2)
it returns  NULL and it should be 20 

select dbo.L_SPLIT('1|20|3|343|44|6|8765',3,'|',3)
It returns  |3|343|44|6|8765  it should be 3|343|44|6|8765

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Stacey Fontenot

Author

Commented:
I am trying to break a string into a number of sections base on a delimiter and return the section of my choice
HainKurtSr. System Analyst

Commented:
ok but the examples you gave does not make sense to me...

for example this
select dbo.L_SPLIT('1|20|3|343|44|6|8765',3,'|',3)
It returns  |3|343|44|6|8765  it should be 3|343|44|6|8765

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Co-Founder and Chief Architect
Top Expert 2016
Commented:
I think I see what you're going for.  Something like:

ALTER FUNCTION [dbo].[L_SPLIT]
(
      @p1 varchar(max), -- string to parse
      @p2 int,      -- number of sections
      @p3 varchar(10),-- Delimiter
	  @p4 int --section to return
)
RETURNS varchar(max)
AS
BEGIN
      -- Declare the return variable here     
	DECLARE @i INT, @result VARCHAR(MAX)
	SET @i = 1

	WHILE @i <= @p2
	BEGIN

		DECLARE @string1 VARCHAR(MAX), @length INT
		IF @i != @p2
		BEGIN
			SET @string1 = (SELECT SUBSTRING(@p1,0,CHARINDEX(@p3,@p1)))
			SET @length = LEN(@string1)
			SET @p1 = (SELECT (SUBSTRING(@p1,@length+2,LEN(@p1))))
		END
		ELSE
		BEGIN
			SET @string1 = @p1
		END

		IF @i = @p4
		BEGIN
			SET @result = @string1
		END

		SET @i = @i + 1

	END

	RETURN @result
END

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Returns:

'1', '20', '3|343|44|6|8765'
sqlresults.png
SharathData Engineer

Commented:
<<
when I run select
          dbo.L_SPLIT('1|20|3|343|44|6|8765',3,'|',3)
It returns  |3|343|44|6|8765  it should be 3|343|44|6|8765 >>
Why it should return 3|343|44|6|8765 and why not just 3?

What is the purpose of 2nd parameter (i.e. sections) in your string?

This function splits the string on the passed delimiter and returns the value at desired location. 
CREATE FUNCTION splitString
(@p1 varchar(max),
 @p2 varchar(1),
 @p3 int
)
RETURNS VARCHAR(100) AS 
BEGIN

 DECLARE @string VARCHAR(100)
 
;WITH ToXml AS
    (
        SELECT  CAST(('<td><![CDATA[' + REPLACE(@p1, @p2, ']]></td><td><![CDATA[') + ']]></td>') AS XML) AS XmlData
    ),
   CTE AS (
    SELECT  td.value('.', 'NVARCHAR(255)') AS [value],
            ROW_NUMBER() OVER (ORDER BY (SELECT 1)) rn
    FROM    ToXml
        CROSS APPLY XmlData.nodes('td') A ( td ))
 SELECT @string = value FROM CTE WHERE rn = @p3
 RETURN @string
 END
 GO
 
 SELECT dbo.splitString('1|20|3|343|44|6|8765','|',1); --1
 SELECT dbo.splitString('1|20|3|343|44|6|8765','|',2); --20
 SELECT dbo.splitString('1|20|3|343|44|6|8765','|',3); --3

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