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RE: [xsl] Get external xml-data thru xsl
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Subject: RE: [xsl] Get external xml-data thru xsl From: "Michael Kay" <mhk@xxxxxxxxx> Date: Thu, 22 Jan 2004 13:05:30 -0000 |
> because of some stupid circumstances I need to grab a RDF -
> File (some sort of a newsticker) and translate it per XSL. I
> am a bloody beginner, so I tried
>
> <xsl:value-of select= "http://theComputer:8080/newsticker.rdf" />
>
The value of the select expression must be an XPath expression, not a
URI. Perhaps you wanted select="document('http://.....rdf')"
But xsl:value-of will flatten the document, which isn't likely to be
what you want.
I would suggest that instead of leaping in with questions like this, you
should spend a week learning the basics of the language.
Michael Kay
> to get the content from the RDF File. But it only gives an
> Error about
>
> ERROR: Description: Expected token 'EOF' found ':'.
>
> http-->:<--//w8r00298:8080/newsticker.rdf
>
> Ok, now I've read, that I can only use Nodes in the select
> statement, so I made a xml:
>
>
> <?xml version="1.0" encoding="ISO-8859-1"?>
> <?xml-stylesheet type="text/xsl" href="testurl.xsl" ?>
>
> <test>
> <uri>
> http://theComputer:8080/newsticker.rdf
> </uri>
> </test>
>
> and a XSL
>
> <?xml version="1.0" encoding="UTF-8"?>
> <xsl:stylesheet
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
> <xsl:template match="/">
> <xsl:variable name="contents_of_uri" select="test/uri" />
> <html>
> <p>
> <xsl:value-of select= "$contents_of_uri" />
> </p>
> </html>
> </xsl:template>
> </xsl:stylesheet>
>
>
> But, as you surely know, this one puts only the URL out. Is
> there a way to get the content of the RDF - File and do some
> translations with it!? Or is this the absolutely wrong way?
>
> My main idea is to get this rdf and format it in some way.
>
>
> Best regards
>
>
> -
> T-Systems
>
>
>
> 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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