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Subject: [xsl] Get external xml-data thru xsl From: "Maasland, Tobias" <Tobias.Maasland@xxxxxxxxxxxxx> Date: Thu, 22 Jan 2004 10:00:32 +0100 |
Hello, 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" /> 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
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