Reversing nodeset

sudhish.sikhamani · June 19th, 2008, 08:15 AM

Hi,

I have a nodeset like given below
<d1>12</d1>
<d1>dfd</d1>
<d1>tyy</d1>
<d1>99</d1>
<d1>klgh</d1>

i would like to reverse it like
<d1>klgh</d1>
<d1>99</d1>
<d1>tyy</d1>
<d1>dfd</d1>
<d1>12</d1>

the number of elements in nodeset can be dynamic.

How can i do it using xslt?

Any help in this regards will be greatly appreciated..

Thanks in advance,
sudhish

Martin Honnen · June 19th, 2008, 08:24 AM

You can process the elements in reverse order using <xsl:sort select="position()"/>:

Code:

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

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

  <xsl:template match="root">
    <xsl:copy>
      <xsl:apply-templates select="d1">
        <xsl:sort select="position()" data-type="number" order="descending"/>
      </xsl:apply-templates>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="d1">
    <xsl:copy-of select="."/>
  </xsl:template>

</xsl:stylesheet>

--
Martin Honnen
Microsoft MVP - XML

sudhish.sikhamani · June 19th, 2008, 11:07 AM

Thank You Martin. That is great. I never thought we can use xsl:sort for this purpose. Thank You Once Again.

mhkay · June 19th, 2008, 11:43 AM

Or in 2.0, use the reverse() function:

<xsl:apply-templates select="reverse(d1)">...

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference