Hi Martin,
Thank you for the quick solution.
Sen
On Sat, Feb 5, 2011 at 6:20 AM, Martin Honnen
<Martin(_dot_)Honnen(_at_)gmx(_dot_)de> wrote:
Senthilukvelaan wrote:
"sections" element is not showing up in my output. I do not know , how
could I include that in my xslt.
Sorry, I did not notice.
You need to write a template for the parent element of "section" elements
then e.g.
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">
<xsl:param name="n" select="2"/>
<xsl:output method="xml" omit-xml-declaration="yes" indent="yes"
encoding="UTF-8" />
<xsl:strip-space elements="*"/>
<xsl:template match="node() | @*">
<xsl:copy>
<xsl:apply-templates select="node()| @*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="*[section]">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates select="section[1]/preceding-sibling::node()"/>
<sections>
<xsl:apply-templates select="section[position() < $n]"/>
</sections>
<xsl:apply-templates select="section[last()]/following-sibling::node()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
--
Martin Honnen
http://msmvps.com/blogs/martin_honnen/
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