Re: testing for last node in a list

[Ann Marie Rubin - Sun PC Networking Engineering <Annmarie dot Rubin at east dot sun dot com>]

Jeni,

Thanks.  It worked!

Ann Marie
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	From: Jeni Tennison <Jeni.Tennison@epistemics.co.uk>
	Subject: Re: testing for last node in a list
	Cc: xsl-list@mulberrytech.com
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	Ann,
	
	>I am generating a list of ancestor nodes for a matched CLASS element. 
The
	XSL 
	>calls this template to output the ancestors when a CLASS is matched. I
	want to 
	>output a "|" character after each CLASS node, EXCEPT the last one.  I 
am
	unable 
	>to express the correct test for the last node in this list. I tried 
using 
	><xsl:if test="position()=last()">, but this statement returns true each
	time the 
	>template is called.
	>
	>Is there another way to solve this?
	
	Try thinking in terms of 'if I just know about node X, what is is about
	node X that affects whether or not there is a | above or below me?'  The
	answer is that all nodes have a | above them apart from the one at the 
top
	of the hierarchy.  How can you tell if you are at the top of the 
hierarchy?
	 Because you don't have a parent node.  So, try:
	
	<!-- named template to do the hierarchy tracing -->
	<xsl:template match="CLASS" mode="hierarchy">
	  <xsl:if test="@SUPERCLASS">
	    <xsl:apply-templates select="key('classes', @SUPERCLASS)"
	mode="hierarchy"/>
	    <xsl:text>|</xsl:text>
	  </xsl:if>
	  <br data="{@SUPERCLASS}">
	    <a href="{@NAME}.html"><xsl:value-of select="@NAME"/></a>
	  </br>
	</xsl:template>
	
	Hope that helps,
	
	Jeni
	
	Dr Jeni Tennison
	Epistemics Ltd, Strelley Hall, Nottingham, NG8 6PE
	Telephone 0115 9061301 • Fax 0115 9061304 • Email
	jeni.tennison@epistemics.co.uk
	
	




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