Subject: RE: new increment question
From: "Michael Kay" <mike@xxxxxxxxxxxx>
Date: Tue, 3 Oct 2006 08:50:26 +0100
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There's a vast amount of flexibility in xsl:number, it's worth reading up on
the instruction. You can achieve this kind of thing quite easily by writing
count="e|f|g".
Michael Kay
http://www.saxonica.com/
> -----Original Message-----
> From: Jonathan Marenus [mailto:jonathanmarenus@xxxxxxxxx]
> Sent: 03 October 2006 03:32
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: new increment question
>
>
> Michael's previous answer helped a lot. Thanks.
>
> The functionality I would like to achieve now is similar to
> the first. The updated XML structure is the following:
>
> <a>
> <b>
> <c>
> <d>
> <e/>
> <f/>
> <g/>
> </d>
> </c>
> <c>
> <d>
> <e/>
> <f/>
> <g/>
> </d>
> </c>
> </b>
> <b>
> <c>
> <d>
> <e/>
> <f/>
> <g/>
> </d>
> </c>
> <c>
> <d>
> <e/>
> <f/>
> <g/>
> </d>
> </c>
> </b>
> </a>
>
> Here, I would like to output an incremental value for each e,f, or g.
>
> In the case above,
>
> first e: 1
> first f: 2
> first g: 3
> second e: 4
> second f: 5
> second g; 6
>
> There will be situations where not all three elements
> (e,f,g) are within d.
>
> So if there is only e in the first d and e,g in the second d,
> the desired output will be:
>
> first e: 1
> second e: 2
> g: 3
>
> The output value should reset to 1 for every new a encountered.
>
> Thank you.
>
> Jonathan Marenus
>
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