Hi Martin,
Thank you for the quick solution.

Sen

On Sat, Feb 5, 2011 at 6:20 AM, Martin Honnen <Martin.Honnen@xxxxxx> 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() &lt; $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/