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Re: [xsl] summing, ok
Subject: Re: [xsl] summing, ok From: Mukul Gandhi <mukul_gandhi@xxxxxxxxx> Date: Sun, 13 Feb 2005 01:41:24 -0800 (PST) |
So you wish to count unique models.. This is a grouping problem... (and Muenchian grouping technique first comes to my mind) Please try this XSL - <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="text" /> <xsl:key name="by-model" match="car" use="model" /> <xsl:template match="/cars"> <xsl:variable name="modelcount"> <xsl:for-each select="car[generate-id(.) = generate-id(key('by-model',model)[1])]"> 1 </xsl:for-each> </xsl:variable> Total no of models - <xsl:value-of select="string-length($modelcount) - string-length(translate($modelcount,'1',''))" /> </xsl:template> </xsl:stylesheet> Regards, Mukul --- Marcos Hercules dos Santos <mhercules@xxxxxxxxx> wrote: > Thanks Mukul Gandhi, > the stage 1 from summing used by exemplification > - cars. It's all ok, cool. > thanks > > Going on, based in the same question, how must do > I to proceed to > group a tag and totalize itself. > > Using the same example: > > <cars> > <car> > <model>V667320</model> > <name>Sportage</name> > <categ>sport</categ> > > </car> > <car> > <model>M382932</model> > <name>Silverado</name> > <categ>pick-up</categ> > </car> > > Imagine that the first element (model) be written > five times, the > second model two times, and other any model four > times. Supposing > that this could be a greatest structure with 25 > models and many > occurences to it , how to do to generate the total > number of models > and not occurences > > The Total number to display = 25 > > if consider the three , then = 3 and not > 5+2+4 > > > Marcos Hercules dos Santos __________________________________ Do you Yahoo!? Yahoo! Mail - Find what you need with new enhanced search. http://info.mail.yahoo.com/mail_250
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