Re: xpath - how to return all nodes but the node match

<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>

best,
-Rob

Hello,

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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