Question

In order to use xpath-functions (specifically the fn part), I included the respective namespace into my xslt stylesheet, like so:

<xsl:stylesheet
  version="2.0"
  xmlns="http://www.w3.org/1999/xhtml"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:fn="http://www.w3.org/2005/xpath-functions"
>

As specified by W3C.

However, when I use fn:document-uri, my XSLT engines tell me I called an unknown function/extension:

  <xsl:variable name="uri" select="fn:document-uri()" />

Opera says:

This document had an invalid XSLT stylesheet. Error message from the XSLT engine:
Error: XPath expression compilation failed: fn:document-uri()
Details: compilation error (characters 1-17, “fn:document-uri()”): unknown function called: ‘{ http://www.w3.org/2005/xpath-functions, document-uri }’

Firefox says:

Error during XSLT transformation: An unknown XPath extension function was called.

And xsltproc refuses transformation, because of xslt 2.0.

So, how do I specify the fn namespace properly?

Answer 1

The problem is that you are using an XSLT 1.0 processor and an XSLT 1.0 processor doesn’t know (and must not know) anything about XPath 2.0 functions.

If you use a real XSLT 2.0 processor, you don’t even have to specify the function namespace — it is a default namespace for any unprefixed function name.

For example, this XSLT 2.0 transformation:

<xsl:stylesheet version="2.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output method="text"/>

 <xsl:template match="/">
    <xsl:sequence select="document-uri(.)" />
    <xsl:text>&#xA;</xsl:text>
    <xsl:sequence select="document-uri(document(''))" />
 </xsl:template>
</xsl:stylesheet>

when executed with Saxon 9.1.5 under the XSelerator, produces correctly the URLs of the source XML document and the stylesheet itself:

file:/C:/Program%20Files/Java/jre6/bin/marrowtr.xml
file:/C:/Program%20Files/Java/jre6/bin/marrowtr.xsl