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position() of a current node

Subject: position() of a current node
From: Daniel Bauke <bonkey@xxxxxxxxxxxxxxxxxxxxx>
Date: Fri, 15 Jun 2001 14:40:37 +0200
daniel bauke
i've checked a faq, but i'm still confused.
i've got:

xsl:
<xsl:template match="chapter">
   <xsl:for-each select="section">
      <xsl:value-of select="position()"/>
   </xsl:for-each>
   <xsl:apply-templates select="section"/>
</xsl:template>

<xsl:template match="section">
   <xsl:value-of select="position()">
</xsl:template>

and an xml like that:
<document>
   <part name="one">
      <section name="1">blah</section>
      <section name="2">bleh</section>
      <section name="3">blah</section>
   </part>
   <part name="two">
      <section name="1">blah</section>
      <section name="2">bleh</section>
      <section name="3">blah</section>
   </part>
</document>

and i'm wondering how to return real position of each section
(now i'm geting 1,3,5)

-- 
Daniel `bonkey' Bauke; http://www.oho.pl/~bonkey/; {happiness=bike&&unix;}

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


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