Re: Counting preceding nodes

Hi Billie,
  Probably the following code might help ..

<xsl:template match="diagram">
    <xsl:attribute name="Number">
        <xsl:value-of select="count(preceding::node()
| ancestor::node())"/>
    </xsl:attribute>		
</xsl:template>

This is not tested. 

(I am relying on this technique based on a problem I
solved earlier.)

Regards,
Mukul

--- Billie <whynot77@xxxxxxxxxxxxx> wrote:

> I need help with a problem I'm having.  I need to
> count the number of nodes that come before the 
> given node, anywhere in the tree.
> 
> So for example, say I have this tree:
> 
> <book>
>     <chapter>
>         <para>
>             Blah blah blah
>             <diagram name="first"/>
>         </para>
>         <para>Blah blah blah</para>
>         <para>
>             <diagram name="second"/>
>             Blah blah blah
>         </para>
>         <diagram name="third"/>
>     </chapter>
>     <chapter>
>         <para>Blah blah blah</para>
>     </chapter>
>     <chapter>
>         <diagram name="fourth"/>
>         <para>Blah blah blah</para>
>     </chapter>
>     <chapter>
>         <para>
>             <quote>
>                 <diagram name="fifth"/>
>                 <diagram name="sixth"/>
>                 Blah blah blah
>             </quote>
>         </para>
>     </chapter>
> </book>
> 
> I gave the diagrams the appropriate names so you
> could see what I was looking for, but basically, 
> I'm looking for a way to number the diagrams based
> on the absolute order in which they are placed.  
> I've tried all sorts of combinations but nothing
> seemed to work.  One logical structure I came up 
> with, that I couldn't figure out how to translate
> into XSL, was: "count the number of diagrams that 
> have an ancestor-or-self which is a
> preceding-sibling to an ancestor-or-self of the
> diagram in 
> question".  Is that possible to do?
> 
> <xsl:template match="diagram">
>     <xsl:attribute name="Number">
>         <xsl:value-of select="???"/>
>     </xsl:attribute>		
> </xsl:template>
> 
> Thanks for your help,
> Billie
> 
> 



		
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