Re: [xsl] Finding the position of node in foreign node list

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


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