Subject: Re: [xsl] Two versions of sum over node list by recursion--why and how does second one work? From: "andrew welch" <andrew.j.welch@xxxxxxxxx> Date: Tue, 5 Sep 2006 17:15:10 +0100 |
SECOND EXAMPLE:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:template name="total-sales-value"> <xsl:param name="list"/> <xsl:choose> <xsl:when test="$list"> <xsl:variable name="first" select="$list[1]"/> <xsl:variable name="total-of-rest"> <xsl:call-template name="total-sales-value"> <xsl:with-param name="list" select="$list[position()!=1]"/> </xsl:call-template> </xsl:variable> <xsl:value-of select="$first/sales * $first/price + $total-of-rest"/> </xsl:when> <xsl:otherwise>0</xsl:otherwise> </xsl:choose> </xsl:template> QUESTION:
In Mr. Kay's example, why does the variable $total-of-rest indeed contain the total and not just the result of the addition in the last recursion?
It's as the recursion unravels - it essentially adds it from last item to the first. It's a great example of a mind-bending recursive programming :)
cheers andrew
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