position() does not return the position of a selected node in the tree, it
returns the position of the current node within the list of nodes that are
currently being processed. You want either count() or xsl:number.
The simplest solution here is XSLT 2.0:
<xsl:number select="leaf[(_at_)attrib='true'][1]"/>
The equivalent in 1.0 is:
<xsl:for-each select="leaf[(_at_)attrib='true'][1]">
<xsl:number/>
</xsl:for-each>
Or you can use count():
count(leaf[(_at_)attrib='true'][1]/preceding-sibling::leaf)+1
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Julien Flotté [mailto:julien_flotte3(_at_)hotmail(_dot_)com]
Sent: 08 June 2007 10:09
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Getting first child node postion with condition
Hi,
I'm a french student in training course.
I make development in xslt and I encounter the following problem :
My xml file :
<?xml version="1.0" encoding="UTF-8"?>
<tree>
<leaf attrib = 'false'>Leaf1</leaf>
<leaf attrib = 'true'>Leaf2</leaf>
<leaf attrib = 'true'>Leaf3</leaf>
</tree>
When I'm in the "tree" template, I want to get the position
of the first child node with the attribute "attrib" equals
"true" but I don't succeed.
I attempt to use the postion function but it doesn't work :
leaf[(_at_)attrib = 'true'][postion()] returns the value of the first child
(Leaf2) but not its position.
Is there a way to to that correctly ?
Cordially,
Julien Flotté.
_________________________________________________________________
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