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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Get external xml-data thru xsl
 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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