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Re: [xsl] Re: Finding the maximum depth from a node
Subject: Re: [xsl] Re: Finding the maximum depth from a node
From: Greg Faron <gfaron@xxxxxxxxxxxxxxxxxx>
Date: Fri, 10 May 2002 14:42:04 -0600
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At 02:11 PM 5/10/2002, you wrote:
Here's a very simple solution -- use FXSL and its maximum()
function/template. Pass to it the list of "folder" nodes and a
reference to your comparison function, which will return 1 if the
"name" attribute of the first argument is bigger than the name
attribute of the second argument.
<xsl:if test="$arg1/@name > $arg2/@name">1</xsl:if>
This only compares the name attribute of the different nodes and returns
that which has the highest value. The original question was to return the
child node that had the largest hierarchy length. The OP might have
confused the issue by making them the same node.
I was thinking something like the following, but it depends on knowing
the length.
<xsl:value-of select="*//*[count(ancestor::*)>3]/@name"/>
Too bad there's not an extension function that works like the following
invalid XSLT code:
<xsl:value-of select="*//*[max(count(ancestor::*))]/@name"/>
If I think of anything, I'll post it.
Greg Faron
Integre Technical Publishing Co.
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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