Subject: RE: [position node] I to get it.
From: "Sergiu Ignat" <sergiu@xxxxxxxxxx>
Date: Wed, 1 Oct 2003 13:13:57 +0300
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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@xxxxxxxxxxxx]
> Sent: 1 octombrie 2003 11:35
> To: XSL-List@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [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
>
>
> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
>
>
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