Subject: Re: [xsl] Finding the position of node in foreign node list From: David Carlisle <davidc@xxxxxxxxx> Date: Thu, 3 Mar 2005 20:35:04 GMT |
> Any ideas how I can find out what the position of the current node is in a > foreign node set? (As opposed to the position in the current node set) Note that sets are unordered data structures, so there isn't really any notion of position within a nde set. It is xsl:for-each which sorts the elements in to document order, not an intrinsic property of the node set. (This changes in xslt2 draft which uses ordered sequences rather than sets) If I understand corectly you want to generate your rows with a td for every item in A and just do something if that item is in B so <xsl:for-each select="$A"> <td> <xsl:if test="count($B|.)=count($B)"> <xsl:apply-templates select="."/> </xsl:if> </td> </xsl:for-each> David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________
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