Subject: RE: [xsl] Directory Structure From: "Evan Lenz" <elenz@xxxxxxxxxxx> Date: Wed, 15 Aug 2001 11:15:01 -0700 |
I'm not sure what context you're thinking of here, but if you're using the common practice of an XML config file which contains URIs, you could simply store the URI in a variable or parameter before drilling down into the document with that URI. For example, <files> <file href="foo.xml"/> <file href="bar/foo.xml"/> </files> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="/"> <xsl:for-each select="/files/file/@href"> <xsl:apply-templates select="document(.)/*"> <xsl:with-param name="$uri" select="string(.)"/> </xsl:apply-templates> </xsl:for-each> </xsl:template> </xsl:stylesheet> The bottom line is that, unless you have access to the URIs in XML, there's no (standard) way of getting them. Hope this helps, Evan Lenz XYZFind Corp. > -----Original Message----- > From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx > [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Eric > Schenfeld > Sent: Wednesday, August 15, 2001 10:31 AM > To: XSL-List@xxxxxxxxxxxxxxxxxxxxxx > Subject: [xsl] Directory Structure > > > Is there a command (or another way that anyone knows of) to find > out the current working directory for a document? So that after > a call, it might return the value: > "/About/LA/Downtown/Restaurants/"? This could be used in a > navigation kind of way as on Yahoo!. > > Thanks for the help. > > eric > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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