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Re: testing for last node in a list
- To: Annmarie dot Rubin at East dot Sun dot COM
- Subject: Re: testing for last node in a list
- From: Jeni Tennison <Jeni dot Tennison at epistemics dot co dot uk>
- Date: Tue, 23 May 2000 10:37:01 +0100
- Cc: xsl-list at mulberrytech dot com
- Reply-To: xsl-list at mulberrytech dot com
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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