Re: [xsl] position within same nodes

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

Larry,

Try (count(preceding-sibling::meep)+1), or 
(count(preceding-sibling::*[name()=name(current())])+1) for the general case.

<xsl:number/> is another way to go. It can get fairly complex (look it up 
in the spec), but in this case you shouldn't need any attributes to 
configure the counting: it'll default to what you want. Bind it to a 
variable if you need to retrieve it through XPath.

Cheers,
Wendell

At 02:35 PM 7/19/01, you wrote:
> What is the XPath to get the position of a given node relative to its
> siblings of the same name?  Eg,
>
> <blah>
>  <argh/>
>  <meep/>
>  <meep/>
>  <stuff/>
>  <meep/>
> </blah>
>
> I'm looking for an XPath statement that would return 1 for the first meep,
> 2 for the second, and 3 for the 3rd meep, rather than 2, 3, and 5 (which
> is what plain position() returns).
>
> I'm trying to count the number of siblings and subtract the number of
> siblings of the same type, then subtract that from position(), but I can't
> get either of the first two parts there to work either.  Does anyone have
> any suggestions?
>
> --Larry Garfield
> lgarfiel@xxxxxxxxxxxxxxxxxxx


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