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RE: accessing last element of node set passed as parameter
- From: "paul morgan" <pmorg at lycos dot com>
- To: xsl-list at lists dot mulberrytech dot com
- Date: Tue, 21 May 2002 06:50:57 -0700
- Subject: RE: [xsl] accessing last element of node set passed as parameter
- Organization: Lycos Mail (http://www.mail.lycos.com:80)
- Reply-to: xsl-list at lists dot mulberrytech dot com
[Kay]: ($result/BAR)[last()]
[] has higher precedence than /
Thanks, I hadn't even thought about operator precedences.
[Pietschmann]: Try select="$result[last()]"
This also seemed to work, judging by the fact that
<xsl:copy-of select="($result/BAR)[last()]" />
and
<xsl:copy-of select="$result[last()]" />
produced equivalent (looking) output. But there was a difference when trying to access the "i" attribute. That is, with the input:
<BAR i=1/><BAR i=2/><BAR i=3/>
and using Michael's solution:
<xsl:variable name="prior" select="($result/BAR)[last()]" />
I need to use the following to access the value of "i"
<xsl:value-of select="$prior/@i" />
but using J's solution:
<xsl:variable name="prior" select="$result[last()]" />
I need to use
<xsl:value-of select="$prior/TEXT/@i" />
Is there a "root" element here that J's solution creates?
Thanks,
Paul
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