[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] 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 http://www.saxonica.com/
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