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Subject: Website+XSLT=source filename?
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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