Re: [xsl] Position of a result tree fragment

Subject: Re: [xsl] Position of a result tree fragment
From: David Carlisle <davidc@xxxxxxxxx>
Date: Thu, 19 Jul 2001 08:58:06 +0100
> If I am passing a result tree fragment from one template to another using
> a variable, is there any way to determine what the fragment's position
> within its parent was originally? 

A result tree fragment isn't part of the original tree it is a
completely new tree with its own root node. If you are trying to
pass a subtree of the source tree then do that (ie use the select
attribute of with-param) then you have access to the source nodes and
can make any query on them, find their parents, count their siblings,


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