Subject: Re: [xsl] grouping list items by attribute From: Jon Gorman <jonathan.gorman@xxxxxxxxx> Date: Wed, 5 Oct 2005 09:18:59 -0500 |
Ack, hit the send button a little early, sorry about that Lynn. For some reason Gmail occassionally focuses on the Send button and I don't quite realize it. In any case, my more complete code "hint" > ie something like (not tested, and just here to provide some hints ) > > <xsl:template name="processList"> > <xsl:param name="nodes" /> > <!-- some of your other stuff and probably need a for loop that > iterates over all the nodes--> > > <xsl:choose> > <xsl:when test="@level >= following-sibling::*[1]/@level"> > <xsl:apply-templates > > select="$nodes/tx.li[1]" mode="in-list"/> > <xsl:call-template name="processList"> <xsl:with-param name="nodes" select="$nodes/tx.li[posiition() > 1]"/> </xsl:call-template> > </xsl:when> > <xsl:when test="@level < following-sibling::*[1]/@level"> > <xsl:apply-templates > select="tx.li[1]" /> <!-- use a variable here to get the value of the position of the next list item that is of the same level as our current one --> <xsl:call-template name="new-list"> <xsl:with-param name="$nodes/tx.li[posiiton() < $nextCurrent and position() > 1]"/> </xsl:call-template> <xsl:call-template name="processList"> <xsl:with-param name="nodes" select="$nodes/tx.li[posiition() = $nextCurrent]"/> </xsl:call-template> Sorry about the email snaffu. Of course, I don't know how great a help my hints are. Jon Gorman
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