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RE: [xsl] [position node] I to get it.
Subject: RE: [xsl] [position node] I to get it. From: "Sergiu Ignat" <sergiu@xxxxxxxxxx> Date: Wed, 1 Oct 2003 13:13:57 +0300 |
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: [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 > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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RE: [xsl] [position node] I to get , Jarno . Elovirta | Thread | RE: [xsl] [position node] I to get , Lionel Crine |
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