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RE: accessing last element of node set passed as parameter
- From: "Michael Kay" <michael dot h dot kay at ntlworld dot com>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Tue, 21 May 2002 08:50:20 +0100
- Subject: RE: [xsl] accessing last element of node set passed as parameter
- Reply-to: xsl-list at lists dot mulberrytech dot com
> I'm trying to figure out how to access the last element of a
> node set which is passed as a parameter to a template.
($result/BAR)[last()]
[] has higher precedence than /
Michael Kay
Software AG
home: Michael.H.Kay@ntlworld.com
work: Michael.Kay@softwareag.com
>
> For example, with:
>
> <xsl:template match="FOO">
> <xsl:param name="result" select="/.." />
> <xsl:variable name="prior" select="$result/BAR[last()]" />
> <!-- Something done it is -->
> </xsl:template>
>
> if "FOO" is matched, and the parameter, "result", is assigned
> the node set
>
> <BAR i=1/><BAR i=2/><BAR i=3/>
>
> what expression would I need to access the last "BAR" element
> (note, I will not always know that the last element has the
> attribute, 'i' equal to 3). I tried many things and I
> thought "$result/BAR[last()]" might be a winner, but (alas) it is not.
>
> Thanks,
>
> Paul
>
>
>
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