Subject: Re: a new(?) grouping problem From: "Nikolai Grigoriev" <grig@xxxxxxxxxxx> Date: Thu, 29 Jun 2000 19:34:21 +0400 |
Mike, > I want to group consecutive days with the same hours together, and just > print the first and last day in each group. The same solution as posted before, but slightly shorter ;-) <?xml version="1.0"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output method="html" version="4.0"/> <!-- Root template: just create a table. --> <xsl:template match="Hours"> <table><xsl:apply-templates/></table> </xsl:template> <!-- Exclude holidays from processing --> <xsl:template match="Holidays" priority="2"/> <!-- Single days, except for holidays. --> <!-- A modeless template creates the row --> <xsl:template match="Hours/*"> <xsl:variable name="hours" select="text()"/> <xsl:if test="not(preceding-sibling::*[not(self::Holidays)][1][text()=$hours])"> <tr> <td> <xsl:value-of select="name()"/> <xsl:apply-templates mode="end" select="following-sibling::*[not(self::Holidays)][1][text()=$hours]"/> </td> <td><xsl:value-of select="."/></td> </tr> </xsl:if> </xsl:template> <!-- A day closes the period if there's no better candidate --> <xsl:template match="Hours/*" mode="end"> <xsl:variable name="hours" select="text()"/> <xsl:choose> <xsl:when test="following-sibling::*[not(self::Holidays)][1][text()=$hours]"> <xsl:apply-templates mode="end" select="following-sibling::*[not(self::Holidays)][1]"/> </xsl:when> <xsl:otherwise> <xsl:text> - </xsl:text><xsl:value-of select="name()"/> </xsl:otherwise> </xsl:choose> </xsl:template> </xsl:stylesheet> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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