RE: [position node] I to get it.

if you match the child document from the parent document you can define a 
parameterized template and call it like this:

<xsl:apply-templates select="dmidref">
                <xsl:with-param name="parentposition" select="position()"/>
</xsl:apply-templates>

and the template for the child must look like this:

<xsl:template match="dmidref">
<xsl:param name="parentposition"/>
        <!-- something -->
        <xsl:value-of select="$parentposition"/>
        <!-- something -->              
</xsl:template>

Sergiu
> -----Original Message-----
> From: Lionel Crine [mailto:crine(_at_)4dconcept(_dot_)fr]
> Sent: 1 octombrie 2003 11:35
> To: XSL-List(_at_)lists(_dot_)mulberrytech(_dot_)com
> Subject: [xsl] [position node] I to get it.
>
>
> Hello,
>
> Id' like to get the position of the parent node of the context node :
>
> For example :
>
> <dmref>
> <dmidref> --> from here I need the position of dmref
> <dmidref>
>
> Is it possible ?
>
> Lionel
>
> Lionel CRINE
> Ingénieur Systèmes documentaires
> Société : 4DConcept
> 22 rue Etienne de Jouy 78353 JOUY EN JOSAS
> Tel : 01.34.58.70.70 Fax : 01.39.58.70.70
>
>
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