Re: [xsl] simulating for with foreach

Subject: Re: [xsl] simulating for with foreach
From: "Mohsen Saboorian" <mohsens@xxxxxxxxx>
Date: Wed, 5 Jul 2006 14:51:04 +0330
Thanks again,

<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? ...

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