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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: How do I change a XSL style sheet to group data t
 Thank you for the information. I have changed it accordingly and nothing seems to happen. I have checked to see if I did everything, checked for any inconsistancies like an extra bracket but still nothing happened. I replace every instance of "sample_date_time" with "concat(sample_date_time, cp_name)". This is what I have done <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output method="xml" indent="yes"/> <!-- the key determines (here) what to group by --> <xsl:key name="sample" match="sample" use="concat(sample_date_time, cp_name)" /> <xsl:template match="/">
<submission imis_company_code="0001843309" ws_name="LENNOX">
<xsl:apply-templates />
</submission>
</xsl:template><xsl:template match="dataroot"> <!-- the actual 'group by' construct --> <xsl:for-each select="sample[generate-id() = generate-id(key('sample', concat(sample_date_time, cp_name)[1])]"> <xsl:copy> <!-- date-time must become an attribute --> <xsl:apply-templates select="concat(sample_date_time, cp_name)" /> <!-- find all samples with equal date time --> <xsl:apply-templates select="key('sample', concat(sample_date_time, cp_name))" /> </xsl:copy> </xsl:for-each> </xsl:template> <!-- match the sample's that are grouped --> <xsl:template match="sample"> <result> <xsl:apply-templates select="*[not(self::concat(sample_date_time, cp_name))]" /> </result> </xsl:template> <!-- any direct child of 'sample' must be turned into an attribute -->
<xsl:template match="sample/*">
<xsl:attribute name="{name()}">
<xsl:value-of select="."/>
</xsl:attribute>
</xsl:template></xsl:stylesheet> Regards. Hennie _________________________________________________________________ Message offline contacts without any fire risk! http://www.communicationevolved.com/en-za/ 
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