RE: [xsl] Directory Structure

["Evan Lenz" <elenz at xyzfind dot com>]

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@lists.mulberrytech.com
> [mailto:owner-xsl-list@lists.mulberrytech.com]On Behalf Of Eric
> Schenfeld
> Sent: Wednesday, August 15, 2001 10:31 AM
> To: XSL-List@lists.mulberrytech.com
> 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
>
>


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