Subject: Re: [xsl] While copying everything, append attribute value with incremental index. From: "Dimitre Novatchev" <dnovatchev@xxxxxxxxx> Date: Sat, 11 Aug 2007 13:01:41 -0700 |
> However, can you explain the logic of the code that you provided: I am afraid you will need to read about the <xsl:number .../> instruction. Generally it is used to "allocate a sequential number to the current node" (from the XSLT 2.0 book of Dr. Kay). -- Cheers, Dimitre Novatchev --------------------------------------- Truly great madness cannot be achieved without significant intelligence. --------------------------------------- To invent, you need a good imagination and a pile of junk ------------------------------------- You've achieved success in your field when you don't know whether what you're doing is work or play On 8/11/07, Xuan Ngo <xuanngo2001@xxxxxxxxx> wrote: > Thank you Dimitre Novatchev for your explanation of the position(). > > However, can you explain the logic of the code that you provided: > <xsl:template match="A[@i]"> > <xsl:variable name="vInum"> > <xsl:number count="A[@i]"/> > </xsl:variable> > <A i="{concat(@i,'_',$vInum)}"> > <xsl:apply-templates select= > "@*[not(name()='i')] > | > node() > " > /> > </A> > </xsl:template> > -------------------------- > I will take a shot here. Tell me if I am right. Each time that it encounters the condition > "A[@i]", it will sum up all preceding "A[@i]". Hence, creating an incremental index(1,2,3,...). > Then, it assigns the sum to the variable vInum so that later it can be used to create the element > A. Since the code is creating the attribute i then it shouldn't copy it. Hence, the condition > "[not(name()='i')]" > > thx! > > > > ____________________________________________________________________________________ > Boardwalk for $500? In 2007? Ha! Play Monopoly Here and Now (it's updated for today's economy) at Yahoo! Games. > http://get.games.yahoo.com/proddesc?gamekey=monopolyherenow
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