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RE: simulating for with foreach

Subject: RE: simulating for with foreach
From: "Michael Kay" <mike@xxxxxxxxxxxx>
Date: Wed, 5 Jul 2006 13:01:32 +0100
RE:  simulating for with foreach
> > <xsl:for-each select="node[position() mod 4 = 1]">
> This is exactly what I want, but although this for-each 
> iterates over 1, 5, 9, ... elements, when I print position() 
> inside this loop, it prints 1, 2, 3, 4, ...
> How can I figutre out which child element is now selected, 1? 5? ...

position() gives you the position of the current node in the current node
list. In the predicate (the thing inside square brackets), the current node
list is "node", that is, all the nodes. In the body of the for-each loop,
the current node list is "node[position() mod 4 = 1]", that is, the list
consisting of every fourth node. Clearly the position of a given node in
these two lists will be different.

If you want, inside the body of the for-each, to know the position of the
node in the original unfiltered list, you will have to do:

<xsl:for-each select="node">
  <xsl:variable name="p" select="position()"/>
  <xsl:if test="$p mod 4 = 1">

Or you could recalculate the position as ((position()-1)*4)+1.

Or if your nodes are siblings, you could use count(preceding-sibling::node),
or xsl:number.

Michael Kay

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