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Re: Getting previous node in nodeset
Subject: Re: Getting previous node in nodeset|
From: Oliver Hallam <oliver@xxxxxxxxxxx>
Date: Fri, 22 Apr 2011 11:18:36 +0100
A slightly neater formulation of this is
$nodelist[subsequence($nodelist,position() + 1,1) is $mynode]
If you know your nodelist is in document order, then the following would
$nodelist[. << $mynode][last()]
On 22/04/2011 10:21, Michael Kay wrote:
On 22/04/2011 08:24, David Carlisle wrote:
On 22/04/2011 02:28, Steve Fogel wrote:
Would appreciate a suggestion:
- I have a node set in the variable $nodelist
- and I have a single node in the variable $mynode
- and the node in $mynode is a member of $nodelist
then in XSLT 2.0, how do I set a variable to contain the node that
is previous to $mynode in $nodelist?
For simplicity and a quick answer, you can assume that all nodes in
$nodelist are siblings, but in reality, $nodelist
contains<topicref>s from a DITA map, so the previous node could be a
sibling, a parent, or the child of the previous sibling.
If I had this problem, I think I would want to take a step back: where
do these two variables come from? Is there any possibility that
instead of setting the variable $mynode to be one of the nodes in
$nodelist, one could set a variable $myNodePosition to be the integer
position of $mynode in $nodelist?
However, for the problem as stated, another option is
$nodelist[(1 to count($nodelist))[subsequence($nodelist, ., 1) is
$mynode] - 1]
In 3.0 this is a classic case for some useful higher-order functions.
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