Re: [xsl] Finding deepest node

Subject: Re: [xsl] Finding deepest node
From: marcus <m-lists@xxxxxxxxxx>
Date: Fri, 02 Dec 2005 21:37:21 +0100
Jon Gorman wrote:
> Well......what do you mean by farthest away? I'm sure it's probably possible.

When I say farthest away I mean the node that has the most nodes between itself and the context node in a straight line (not counting siblings).

.//node[@a = 'avalue' and max(count(ancestor::*))]/@id

I know that isn't a valid expression but I hope you know what I mean


 <node a="1" id="1"/>
 <node a="2" id="2">
  <node a="1" id="3">
   <node a="1" id="3.1"/>
  <node a="1" id="4"/>
 <node a="2" id="5">
  <node a="1" id="6">
   <node a="1" id="7"/>

Say that a node must have @a=1 to be selectable.

In this example the nodes with @id=3.1 and @id=7 are furthest away from <nodes> and have a distance of 3. @id=3, @id=4, @id=6 have a distance of 2 and so on.


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