<xsl:variable name="path" select="concat(?PATHINFOMISSING?,
@href)"/>
<xsl:variable name="external-doc" select="document($path)"/>
Is there a way to get ?PATHINFOMISSING? / the path of the transformed
document?
In XPath 2.0 you can get the base URI of a node using the base-uri()
function. Note that the way to combine a base URI with a relative URI is not
using concat(), but using the resolve-uri() function - the algorithm is in
RFC 2396, and is only the same as concatenation in a few special cases.
Michael Kay
http://www.saxonica.com/
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