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Stacey Fontenot
 asked on

section a string

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
Microsoft SQL Server

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Last Comment
Sharath S

8/22/2022 - Mon
HainKurt

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

ASKER
I am trying to break a string into a number of sections base on a delimiter and return the section of my choice
HainKurt

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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Sharath S

<<
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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