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How does one go about finding out the XML source filename in XSL? I tried to utilise the document-uri function of XSL 2.0; but am getting an error (probably from the xsltproc processor) about only being version 1.0 capable. I currently have successfully been using the following: <snip> <xsl:variable name="page" select="."/> <xsl:variable name="xml_source"> <xsl:choose> <xsl:when test="$page/config[@param='source']"> <xsl:value-of select="($page/config[@param='source'])[1]/@value"/> </xsl:when> <xsl:otherwise> <xsl:text>index.xml</xsl:text> </xsl:otherwise> </xsl:choose> </xsl:variable> </snip> As you can see, I am having to generate a specific element to hold the source filename. But, given the volume of pages that I am constructing(4,000+); I would rather utilise a construct during transformation so that I don't need to alter the XML sources and add this element. Thanks, Thomas
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