I am looking to return the content of a particular XML Tag <para> without its sub tags <bridgehead> or <sliceXML> in the results. I am testing the following useing http://xslttest.appspot.com/ Any help is as always, much appreciated.
My XML
<para>
<bridgehead>Galaxy Zoo</bridgehead>
<sliceXML>Galaxy</sliceXML>
The human eye is far better at identifying characteristics of galaxies
than any computer. So Galaxy Zoo has called for everyday citizens to
help in a massive identification project. Well over a hundred thousand
people have helped identify newly discovered galaxies. Now you can, too.
</para>
My XSLT
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:sparql-results="http://www.w3.org/2005/sparql-results#" version="1.0">
<xsl:template match="/">
<xsl:call-template name="results"/>
<xsl:message>FROM simpleHMHTransform XSLT8</xsl:message>
</xsl:template>
<xsl:template name="results">
<xsl:for-each select="//para">
<xsl:call-template name="para"/>
</xsl:for-each>
</xsl:template>
<xsl:template name="para">
<div id="para">
<xsl:value-of select="."/>
</div>
</xsl:template>
</xsl:stylesheet>
My current results
<?xml version="1.0" encoding="UTF-8"?><div xmlns:sparql- results="http://www.w3.org/2005/sparql-results#" id="para">
Galaxy Zoo
Galaxy
The human eye is far better at identifying characteristics of galaxies
than any computer. So Galaxy Zoo has called for everyday citizens to
help in a massive identification project. Well over a hundred thousand
people have helped identify newly discovered galaxies. Now you can, too.
</div>
My desired results
<?xml version="1.0" encoding="UTF-8"?><div xmlns:sparql-results="http://www.w3.org/2005/sparql-results#" id="para">
The human eye is far better at identifying characteristics of galaxies
than any computer. So Galaxy Zoo has called for everyday citizens to
help in a massive identification project. Well over a hundred thousand
people have helped identify newly discovered galaxies. Now you can, too.
</div>
1 Answer