In 2.0:
<xsl:for-each-group select="//title"
group-by="count(ancestor::*)">
<xsl:sort select="current-grouping-key()">
<xsl:if test="position()=last()">
<xsl:copy-of select="current-group()"/>
</
</
In 1.0 you can usually tackle this kind of problem by sorting nodes
according to their depth (that is, count(ancestor::*)) and taking the first
or last in sorted sequence. It's complicated here because you want all the
nodes having that maximum depth. That suggests two passes, one to get the
max depth and the second to select nodes with that depth
<xsl:variable name="max-depth">
<xsl:for-each select="//title">
<xsl:sort select="count(ancestor::*)" data-type="number"/>
<xsl:if test="position()=last()">
<xsl:copy-of select="count(ancestor::*)"/>
</
</
<xsl:copy-of select="//title[count(ancestor::*) = $max-depth]"/>
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Bert [mailto:arm(_at_)xs4all(_dot_)nl]
Sent: 07 April 2005 16:59
To: Xsl-List
Subject: [xsl] How to find the deepest node?
Hi,
I have the following XML-file (see below) and I need to find
the deepest
node of 'title'.
The resultfile should only contain the titles 'Text 02' and 'Text 03',
because they are in this case the deepest nodes. The next
XML-file I receive
may contain more nested levels of items and whatever the
level of nesting I
only need to find the deepest titles.
I know I have to do something with recurse, but I don't know
how to start.
Who can help me on this one?
Kind regards,
Bert
<?xml version="1.0" encoding="UTF-8"?>
<document>
<item>
<item>
<title>Text 01</title>
</item>
<item>
<item>
<title>Text 02</title>
</item>
</item>
</item>
<item>
<item>
<item>
<title>Text 03</title>
</item>
</item>
</item>
<item>
<item>
<title>Text 04</title>
</item>
</item>
</document>
Resultfile should contain:
<?xml version="1.0" encoding="UTF-8"?>
<document>
<item>
<item>
<item>
<title>Text 02</title>
</item>
</item>
</item>
<item>
<item>
<item>
<title>Text 03</title>
</item>
</item>
</item>
</document>
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