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Re: XPath selecting chain of following siblings of the

Subject: Re: XPath selecting chain of following siblings of the same name
From: "Mukul Gandhi" <gandhi.mukul@xxxxxxxxx>
Date: Sun, 11 Mar 2007 10:24:03 +0530
Re:  XPath selecting chain of following siblings of the
Thanks Ken for the answer. Now it's clear to me.

On 3/11/07, G. Ken Holman <gkholman@xxxxxxxxxxxxxxxxxxxx> wrote:
At 2007-03-10 17:13 +0530, Mukul Gandhi wrote:
>I wonder, why this XSLT 1.0 solution doesn't work:

 From what I can tell it is because you only look at the "next" a and
not all contiguous a elements that follow:

> <xsl:apply-templates select="following-sibling::*[1][self::a]" />

Did you mean to do what is in the code below? This appears to work just fine.

I hope this helps.

. . . . . . . . . Ken

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

<xsl:output method="xml" indent="yes"/>

<!--identity transform for all nodes other than 'a'-->
<xsl:template match="node() | @*">
    <xsl:apply-templates select="node() | @*" />

<!--the first a-->
<xsl:template match="a[not(preceding-sibling::*[1][self::a])]">
    <xsl:apply-templates select="." mode="nested"/>

<!--other a's when nested-->
<xsl:template match="a" mode="nested">
  <xsl:copy-of select="."/>
  <xsl:apply-templates select="following-sibling::*[1][self::a]"

<!--other a's when not nested can be ignored-->
<xsl:template match="a"/>


-- World-wide corporate, govt. & user group XML, XSL and UBL training RSS feeds: publicly-available developer resources and training G. Ken Holman mailto:gkholman@xxxxxxxxxxxxxxxxxxxx Crane Softwrights Ltd. http://www.CraneSoftwrights.com/s/ Box 266, Kars, Ontario CANADA K0A-2E0 +1(613)489-0999 (F:-0995) Male Cancer Awareness Aug'05 http://www.CraneSoftwrights.com/s/bc Legal business disclaimers: http://www.CraneSoftwrights.com/legal

Mukul Gandhi

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