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[xsl] Is this node, the document root... best test?


Subject: [xsl] Is this node, the document root... best test?
From: Stephen Cunliffe <scunliffe@xxxxxxxxxxxxxxxxxx>
Date: Mon, 20 Oct 2003 10:58:58 -0400

Hello XSL group;

In a template, matching on "*", I need to test, if this node, is the document root node.

E.g.

<xsl:template match="*" mode="someTemplate">
 <xsl:choose>
   <!-- Method (1) -->
   <xsl:when test="not(parent::*)">
     This works, but is it safe/efficient?
   </xsl:when>
   <!-- Method (2) -->
   <xsl:when test="string(name(parent::*)) = ''">
     This works, but is it safe/efficient?
   </xsl:when>
   <xsl:otherwise>This node, is *NOT* the root node...</xsl:otherwise>
 </xsl:choose>
</xsl:template>

Question:
What is the best/simplest/recommended method (must be non-parser specific)?
In terms of processing speed?
In terms of just plain cool?
(e.g. I was hoping that "not(..)" would work ;-)


I'm open to any/all options, the 2 above are just one's I've had luck with.

PS I'm currently leaning towards Method 1, above.

PPS I want to be careful, that I don't try and match the
   node's "name" against the document root node's "name",
   as this may provide a false match.

Cheers,

Steve



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