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[xsl] is there a way to output XML with invalid qnames?
Subject: [xsl] is there a way to output XML with invalid qnames? From: Jessica Britton <jessica.britton@xxxxxxxxxxxxxxx> Date: Fri, 6 Nov 2009 16:57:46 -0800 |
I'm trying to format XML files so that I can display them as examples in a wiki. I'd like the element names in the examples to link to other wiki pages, so I'm trying to figure out the best way to take source like this: <ns:elementName1> <ns:elementName2>value</ns:elementName2> </ns:elementName1> And get output like this: <[ns:elementName1|page 1]> <[ns:elementName2|page 2]>value</ns:elementName2> </ns:elementName1> Of course, "[ns:elementName|page]" isn't a valid qname so just trying to rename the elements in an XML to XML transform doesn't work. I've messed around a bit trying to output HTML that looks the way I want it to look, including indentation, eg: <xsl:for-each select="rss/channel/title|rss/channel/link|rss/channel/description"> <xsl:variable name="value" select="text()" /> <xsl:variable name="name" select="name()" /> <xsl:variable name="newName" select="concat($open, $name, $linkEnd)" /> <<xsl:value-of select="$newName"/>><xsl:value-of select="$value"/></<xsl:value-of select="$name"/>><br /> </xsl:for-each> That works fine for simple inline elements, but it I'm not sure how to deal with nodes with children. Before I give myself a headache trying to figure it out, I was wondering if anyone could suggest a better approach I should look into.
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