Subject: Re: [xsl] xpath - how to return all nodes but the node matching a value in an arbitrary tree?|
From: Robert Koberg <rob@xxxxxxxxxx>
Date: Sat, 04 May 2002 10:40:00 -0700
<xsl:template match="node()|@*"> <xsl:if test="not(@id=$id)"> <xsl:copy> <xsl:copy-of select="@*"/> <xsl:apply-templates/> </xsl:copy> </xsl:if> </xsl:template>
I have a heirarchical categorization which I'm trying to manipulate via XSL
In order to "delete" a node, I'm trying to write an xpath that, when given an id, will return the entire tree with the exception of the matching node for the id. I've tried many possibilities but can't seem to get the thing to work -- many times it just skips the match all together and returns the whole tree. Since the heirarchical structure is arbitrary, I cannot just pinpoint the exact location of the node to be removed -- I suppose I could with two stylesheets, but I have a feeling there is a more elegant solution with a single xpath.
Can someone please help me with the xpath that would return the whole tree, minus the node?
Thanks, Scott Zagar
Here's a mockup of the XML ... the structure is arbitrary:
<category id="5">music <category id="6">blues <artist id="7">bb king</artist> <artist id="8">lightnin hopkins</artist> </category> <category id="6">jazz <category id="11">bebop <artist id="12">charlie parker</artist> </category> <artist id="9">miles davis</artist> <artist id="10">john coltrane</artist> </category> </category>
Here's the XSL, in this example I'm trying to return everything but the artist node matching the delete_id parameter :
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > <xsl:output method="xml" indent="yes" omit-xml-declaration="yes" cdata-section-elements="category artist" /> <xsl:param name="delete_id"/> <xsl:template match="/"> <xsl:copy-of select="//category//artist[@id!=$delete_id]"/> <--------- this is it </xsl:template> </xsl:stylesheet>
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