From RSS to XML with XSL
Hello again,
Here I want to have to publication elements sorted by category and then by title in XML to be created. Is it possible
[code]
<?xml version="1.0" standalone="yes"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxsl="urn:schemas-microsoft-com:xslt"
xmlns:user="http://www.swissre.com" version="1.0">
<xsl:output method="xml"/>
<xsl:strip-space elements="*" />
<xsl:template match="/">
<rss version="2.0" xmlns:sreportal="http://intrawes.zrh.swissre.com/webapp/epv2">
<channel>
<title></title>
<description>Cards generated by plumtree publication_TEST</description>
<language>en-us</language>
<link>http://intrawes.zrh.swissre.com/webapp/intranet/</link>
<pubDate></pubDate>
<lastBuildDate></lastBuildDate>
<category></category>
<generator></generator>
<docs>http://backend.userland.com/rss</docs>
<cloud></cloud>
<ttl></ttl>
<skipHours></skipHours>
<skipDays></skipDays>
<xsl:apply-templates select="//CARD"/>
</channel>
</rss>
</xsl:template>
<xsl:template match="CARD">
<xsl:text disable-output-escaping="yes"><item></xsl:text>
<xsl:apply-templates select="@*"/>
<xsl:text disable-output-escaping="yes"></item></xsl:text>
</xsl:template>
<xsl:template match="transform">
<xsl:if test="@from='open_document_url'">
</xsl:if>
</xsl:template>
<xsl:template match="@*">
<xsl:param name="thetag" select="name(.)"/>
<xsl:param name="thetagname" select="//transform[@from=$thetag]/@to"/>
<xsl:text disable-output-escaping="yes"><</xsl:text>
<xsl:value-of select="$thetagname"/>
<xsl:text disable-output-escaping="yes">></xsl:text>
<xsl:value-of select="."/>
<xsl:text disable-output-escaping="yes"></</xsl:text>
<xsl:value-of select="$thetagname"/>
<xsl:text disable-output-escaping="yes">></xsl:text>
<xsl:if test="$thetagname='link'">
<xsl:text disable-output-escaping="yes"><guid isPermaLink="true"></xsl:text>
<xsl:value-of select="."/>
<xsl:text disable-output-escaping="yes"></guid></xsl:text>
</xsl:if>
</xsl:template>
</xsl:stylesheet>
[code]
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