Retrieving the position of an ancestor in nested <xsl

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Subject: Retrieving the position of an ancestor in nested <xsl:for-each>
From: "Jonny Pony" <jonnypony666@xxxxxxxxxxx>
Date: Tue, 04 May 2004 10:54:38 +0000

Hi,

how can I retrieve the position of an ancestor in a nested <xsl:for-each>-loop?

My xml:

<?xml version="1.0" encoding="UTF-8"?>
<superNode>
	<Node>
		<SubNode>bla</SubNode>
		<SubNode>bla</SubNode>
		<SubNode>bla</SubNode>
	</Node>
	<Node>
		<SubNode>bla</SubNode>
	</Node>
	<Node>
		<SubNode>bla</SubNode>
		<SubNode>bla</SubNode>
	</Node>
	<Node>
		<SubNode>bla</SubNode>
	</Node>
</superNode>

The nested <xsl:for-each>:

<xsl:for-each select="Node"> ... <xsl:for-each select="SubNode"> Position of Node: <xsl:value-of select="??????????"/>  Position of SubNode: <xsl:value-of select="position()"/> </xsl:for-each> </xsl:for-each>


Thanks
Jonny

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Current Thread
Retrieving the position of an ancestor in nested <xsl:for-each> Jonny Pony - Tue, 04 May 2004 10:54:38 +0000 <= David Carlisle - Tue, 4 May 2004 12:14:21 +0100 <Possible follow-ups> Jarno.Elovirta - Tue, 4 May 2004 14:10:29 +0300

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Re: ancestor axis order, David Carlisle	Thread	Re: Retrieving the position o, David Carlisle
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