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Re: [xsl] XPATH for attribute with different namespace from element
Subject: Re: [xsl] XPATH for attribute with different namespace from element From: Joerg Pietschmann <joerg.pietschmann@xxxxxx> Date: Mon, 04 Feb 2002 09:42:38 +0100 |
Edward.Middleton@xxxxxxxxxxx wrote: > I am trying to make a template match for an rss about attribute > but I can't seem to get anything to work. ... > This is my rss file > <?xml version="1.0" encoding="UTF-8"?> > <rdf:RDF xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#" > xmlns="http://purl.org/rss/1.0"> ^^^^^^^^^^^^ Here is the problem: you are putting the "channel" element in a namespace. This means... > <xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" > xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#" > xmlns="http://purl.org/rss/1.0"> ... > <xsl:template match="rdf:RDF/channel/@rdf:about"> ^^^^ ...in order to match it, is has to be in the namespace too. Declaring the default namespace as you did above only affects literal result elements in the style sheet, not the XPaths araund in matches and selects. You have to use an explicit namespace prefix: <xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#" xmlns:rss="http://purl.org/rss/1.0"> ... <xsl:template match="rdf:RDF/rss:channel/@rdf:about"> That's becoming a FAQ. HTH J.Pietschmann XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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