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(_at_)yahoo(_dot_)com]
Sent: 03 October 2006 03:32
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] 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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