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[xsl] position() of a current node
Subject: [xsl] position() of a current node From: Daniel Bauke <bonkey@xxxxxxxxxxxxxxxxxxxxx> Date: Fri, 15 Jun 2001 14:40:37 +0200 |
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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