[xsl] Position of a Child

Subject: [xsl] Position of a Child
From: "Casadome, Francisco Javier" <Francisco.Casadome@xxxxxxxxxxxxxx>
Date: Thu, 20 Sep 2001 20:17:43 +0200
Hi all,

I'm having problems getting the child's position() of the context node.
I reference the child by its name (stored in a variable), like this:

<xsl:value-of select="*[name()=$NodeName]"/>

What I would like to get is the position of that child inside the context
node.
I don't seem to find the correct syntax for this.

Thanks in advance,
Frank.

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