Subject: RE: [xsl] Grouping problem From: "Joel Dubien" <joel@xxxxxxxxxxxxxxx> Date: Thu, 8 Apr 2010 12:44:33 -0600 |
Thank you very much Martin. Because this grouping uses keys, there would be no way to use a recursive template? I would like to apply this transformation on any XML structure, and pass the entities involved on to the template that transforms it. This may be more work than it is worth, and I could easily code up one of these nested loops for each required entity grouping. I was just wondering if it was possible to do in a more generic / dynamic manner. Thanks again, Joel -----Original Message----- From: Martin Honnen [mailto:Martin.Honnen@xxxxxx] Sent: Thursday, April 08, 2010 12:27 PM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] Grouping problem Joel Dubien wrote: > Desired Output: > <item> > <master_affiliate_id>68800</master_affiliate_id> > <queries_by_period>75,120,0,0</queries_by_period> > <subitems> > <item> > <affiliate_id>68801</affiliate_id> > <queries_by_period>75,120,0,0</queries_by_period> > </item> > </subitems> > </item> > <item> > <master_affiliate_id>69767</master_affiliate_id> > <queries_by_period>75,120,0,0</queries_by_period> > <subitems> > <item> > <affiliate_id>69775</affiliate_id> > <queries_by_period>75,120,0,0</queries_by_period> > </item> > </subitems> > </item> Here is an XSLT 1.0 stylesheet that produces the described output: <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output indent="yes"/> <xsl:param name="plist" select="'test2010040802.xml'"/> <xsl:variable name="periods" select="document($plist)/period_list/period"/> <xsl:variable name="main-root" select="/"/> <xsl:key name="k1" match="master_affiliate" use="master_affiliate_id"/> <xsl:key name="k2" match="master_affiliate/summary" use="concat(../master_affiliate_id, '|', period)"/> <xsl:key name="k3" match="affiliates/affiliate" use="affiliate_id"/> <xsl:key name="k4" match="affiliates/affiliate" use="concat(affiliate_id, '|', period)"/> <xsl:template match="/"> <xsl:for-each select="descendant::master_affiliate[generate-id() = generate-id(key('k1', master_affiliate_id)[1])]"> <xsl:variable name="cgk" select="master_affiliate_id"/> <item> <xsl:copy-of select="master_affiliate_id"/> <queries_by_period> <xsl:for-each select="$periods"> <xsl:variable name="period" select="."/> <xsl:for-each select="$main-root"> <xsl:variable name="s" select="key('k2', concat($cgk, '|', $period))"/> <xsl:choose> <xsl:when test="$s"> <xsl:value-of select="$s/queries"/> </xsl:when> <xsl:otherwise>0</xsl:otherwise> </xsl:choose> </xsl:for-each> <xsl:if test="position() != last()"><xsl:text>,</xsl:text></xsl:if> </xsl:for-each> </queries_by_period> <subitems> <xsl:for-each select="key('k1', master_affiliate_id)/affiliates/affiliate[generate-id() = generate-id(key('k3', affiliate_id)[1])]"> <item> <xsl:copy-of select="affiliate_id"/> <xsl:variable name="cgk1" select="affiliate_id"/> <queries_by_period> <xsl:for-each select="$periods"> <xsl:variable name="period" select="."/> <xsl:for-each select="$main-root"> <xsl:variable name="s" select="key('k4', concat($cgk1, '|', $period))"/> <xsl:choose> <xsl:when test="$s"> <xsl:value-of select="$s/queries"/> </xsl:when> <xsl:otherwise>0</xsl:otherwise> </xsl:choose> </xsl:for-each> <xsl:if test="position() != last()"><xsl:text>,</xsl:text></xsl:if> </xsl:for-each> </queries_by_period> </item> </xsl:for-each> </subitems> </item> </xsl:for-each> </xsl:template> </xsl:stylesheet> I am not quite sure however it will work for more complex input documents as I am not sure the same affiliate could occur under different master_affiliate elements. In that case the key 'k4' would need to include the master affiliate id as well. -- Martin Honnen http://msmvps.com/blogs/martin_honnen/
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