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position() of a current node
- To: xsl-list at lists dot mulberrytech dot com
- Subject: [xsl] position() of a current node
- From: Daniel Bauke <bonkey at sokrates dot mimuw dot edu dot pl>
- Date: Fri, 15 Jun 2001 14:40:37 +0200
- Reply-To: xsl-list at lists dot mulberrytech dot com
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;}
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