Re: [xsl] Accessing nth element

Subject: Re: [xsl] Accessing nth element
From: "bharat chintapally" <hydbad@xxxxxxxxxxx>
Date: Sun, 06 May 2001 16:22:03 -0400
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..


Janning Vygen <vygen@xxxxxxxxxxxx> Sent by: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx 05/01/01 09:23 AM Please respond to xsl-list

		 To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx, Bharat.Chintapally@xxxxxxxxxxxxxxxxx
		 Subject: Re: [xsl] Accessing nth element

Am Dienstag, 1. Mai 2001 15:04 schrieb Bharat.Chintapally@xxxxxxxxxxxxxxxxx:
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

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.


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