Subject: Re: [xsl] position of arbitrary parent From: David Carlisle <davidc@xxxxxxxxx> Date: Tue, 17 Jul 2001 19:05:47 +0100 |
> which leaves me back with needing to know the > position() of the current <chapter> element. Probably you need to pass the chapter node as a parameter down to you html templates. <xsl:for-each select="chapter"> <xsl:apply-templates select="$chapterTemplate/html" > <xsl:with-param name="c" select="."/> <xsl:apply-templates/> </xsl:for-each> and then declare c param in your templates that handle your html doc, David _____________________________________________________________________ This message has been checked for all known viruses by Star Internet delivered through the MessageLabs Virus Scanning Service. For further information visit http://www.star.net.uk/stats.asp or alternatively call Star Internet for details on the Virus Scanning Service. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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