Hi,
I am new to XSL. I need help in passing the below RSS Feed URL to through this XSL Code so that it fetches the Title and description from the Feed:
The URL that i need to pass is :
http://infotech.indiatimes.com/rssfeeds/2663900.cms
Where exactly do i need to include this URL, someone please help, this is an urgent requirement :(
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="
http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html" indent="yes"/>
<xsl:param name="TITLE"/>
<xsl:template match="rss">
<!-- Do not show channel image -->
<xsl:for-each select="channel/item">
<xsl:if test="position() <= 7">
<br>
<strong><a href="{link}" target="_new"><xsl:value-of select="title"/></a></strong><br></br>
<!-- only display markup for description if it's present -->
<xsl:value-of select="description" disable-output-escaping="yes"/>
</br>
<br></br>
</xsl:if>
</xsl:for-each>
</xsl:template>
<xsl:template match="description">
<br>
<xsl:value-of select="."/>
</br>
</xsl:template>
</xsl:stylesheet>