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On 22/04/2011 08:24, David Carlisle wrote:
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.
Re: [xsl] Getting previous node in nodeset
Subject: Re: [xsl] Getting previous node in nodeset From: Michael Kay <mike@xxxxxxxxxxxx> Date: Fri, 22 Apr 2011 10:21:21 +0100 |
On 22/04/2011 08:24, David Carlisle wrote:
On 22/04/2011 02:28, Steve Fogel wrote:Hi, all...
Would appreciate a suggestion:
If:
- 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.
Many thanks
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.
Michael Kay Saxonica
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