Can someone answer this question? I'll paste the full stylesheet at
the bottom.
On Nov 10, 2004, at 5:56 PM, Bruce D'Arcus wrote:
I'm getting this error though:
A sequence of more than one item is not allowed as the second
argument of concat()
Here's how I've defined the top-level variables:
<xsl:variable name="bibkey" select="//db:biblioref/@linkend" />
<xsl:variable name="bibrecord" select="doc(concat('bib-data/',
$bibkey, '.mods'))" />
<xsl:key name="biblio" match="//mods:mods" use="@ID" />
Why am I getting that error then?
Bruce
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="2.0"
xmlns:mods="http://www.loc.gov/mods/v3"
xmlns:db="http://docbook.org/docbook-ng"
exclude-result-prefixes="db mods xs">
<xsl:output method="xhtml" encoding="UTF-8" />
<xsl:variable name="bibkey" select="//db:biblioref/@linkend"/>
<xsl:variable name="bibrecord" select="doc(concat('bib-data/',
$bibkey, '.mods'))" />
<xsl:key name="biblio" match="//mods:mods" use="@ID" />
<xsl:template match="/">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Testing</title>
</head>
<body>
<h3>Citations</h3>
<xsl:for-each select="//$bibkey">
<ol>
<li><xsl:value-of select="." /></li>
</ol>
</xsl:for-each>
<h3>Titles</h3>
<xsl:for-each select="$bibrecord/key('biblio', $bibkey)">
<xsl:value-of select="mods:title" />
</xsl:for-each>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
Source:
<?xml version="1.0" encoding="utf-8"?>
<article xmlns="http://docbook.org/docbook-ng">
<info>
<title>Test</title>
</info>
<section>
<info>
<title>Introduction</title>
</info>
<para>Some citations: <citation><biblioref linkend="one"/><biblioref
linkend="two"/><biblioref linkend="three"/></citation>.</para>
<para>A citation with page number detail: <citation><biblioref
linkend="one"
units="page" begin="23" end="24" /></citation>. A
citation <footnote><para>... in a footnote <citation><biblioref
linkend="three" begin="234"/></citation></para></footnote>.</para>
</section>
<bibliography/>
</article>
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