If position() returns even numbers, it's because you are counting text nodes
as well as elements. The answer is to count only the elements, which will
happen if you use <xsl:for-each select="*"> or <xsl:apply-templates
select="*"> rather than using select="node()".

Dividing position() by 2 is wrong, because the whitespace text nodes won't
always be there.

Alternatively, use xsl:number.

Michael Kay
http://www.saxonica.com/

> -----Original Message-----
> From: Mohsen Saboorian [mailto:mohsens@xxxxxxxxx] 
> Sent: 26 June 2006 12:06
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject:  counting the element number in a recursive tree
> 
> Hi,
> I'm confused with counting the current element number with a 
> recursive xml structure. Here is a sample:
> <node>
>     <node>
>     </node>
>     <node>
>         <node>
>             <node>
>             </node>
>         </node>
>     </node>
>     <node>
>     </node>
> </node>
> 
> = = = = = = = = =
> Since position() returns even numbers (I think is counts also 
> #text elements), I used position div 2, but unfortunately 
> deeper nodes also have problems.
> 
> <xsl:template match="node">
>     <xsl:value-of select="position() div 2" />
> 
>     <xsl:apply-templates />
> </xsl:template>
> 
> What souled I do to find out the correct number of node I'm 
> currently on.
> 
> Thanks in advance.