Re: [xsl] Accessing nth element

["bharat chintapally" <hydbad at hotmail dot com>]

Hi janning:
	 Thanks for your response..But I did not see any difference in performance 
when I tried the following
<xsl:for-each select="./records[position() < 20]">

or when I used template for record.

My question is how does the select in the for-each behave.. When we select 
by using the "select="./records[position() < 20]" does the XSLT processor 
return when the the cursor (db like) moves past
record # 20 (like a for loop; return when the termination condition is met) 
or does it go loop all the n records
(in particular 20 thru n).

The more I look at it, I feel that I should be using a for-loop (or 
recursion) here..

Any suggestions are welcome..

Thanks
--bharat


	Janning Vygen <vygen@planwerk6.de>
	Sent by: owner-xsl-list@lists.mulberrytech.com
	05/01/01 09:23 AM
	Please respond to xsl-list

		 To: xsl-list@lists.mulberrytech.com, Bharat.Chintapally@CommerceQuest.com
		 cc:
		 Subject: Re: [xsl] Accessing nth element


Am Dienstag,  1. Mai 2001 15:04 schrieb 
Bharat.Chintapally@CommerceQuest.com:
>Hello all:
>      I think I need to replace for-each in my XSLT's with recursion. I am
>interested in picking 'x' elements from 'y' (20 for each page from 500
>records) for paging. for-each is working fine, but it is overkill, I should
>be able to terminate (return from as in procedural lang's) processing when
>I finish processing 20 records for a given page.
>      I am trying to switch to recursion, but have a quick question. Is
>there a way to fetch an n'th element from given bunch of records. For
>example when I am processing records 20 thru records 40 out of 500 records,
>I would like directly fetch the record #20 from all the records. Is it
>possible..

try something like this

<xsl:for-each select="./records[position() < 20]">

not tested, but with position function inside your XPath expression you
should get the records you want.

janning

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