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

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Subject: Re: xpath - how to return all nodes but the node matching avalue in an arbitrary tree?
From: Joerg Heinicke <joerg.heinicke@xxxxxx>
Date: Sat, 04 May 2002 20:24:57 +0200

Hello Robert,

I don't think that there are significant performance differences.

Your <xsl:copy-of select="@*"/> shell be a bit faster than my <xsl:apply-templates select="@*"/>, because the template for the attributes must not be searched. I use the apply-templates, because I can easily add a new template e.g. for a special attribute.

And <xsl:apply-templates/> is the same as <xsl:apply-templates select="node()"/>. I use normally the "special" version with "text()|*", because I don't want to copy comments or the xsl-stylesheet processing-instruction.

The last difference is the <xsl:if> and my predicate at <xsl:apply-templates>. It's maybe the same as with the attributes. Normally I would use a simple apply-templates and again a special template: <xsl:template match="*[@id = $delete_id]"/> But this is not possible with variables.

Joerg

Robert Koberg schrieb:

very cool Joerg!

Does this have performance benefits (or negatives) over the (modified) standard identity template:

<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

XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list

Current Thread

xpath - how to return all nodes but the node matching a value in an arbitrary tree?, (continued)
- Joerg Heinicke - Sat, 4 May 2002 09:19:39 -0400 (EDT)
  - Wendell Piez - Mon, 6 May 2002 12:36:12 -0400 (EDT)

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Re: xpath - how to return all, Robert Koberg	Thread	Re: xpath - how to return all, Jeni Tennison
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