Re: position() of the parent

> Worked great! But just for my curiosity, is there any way to find the
> position() of the parent? 

the question is not well formed. Nodes do not, of themselves, have a
position, they only have a position in a node list.

So for example if you select the parent with .. or parent:: the
position() will be 1, as it will be the first node in a list of one, but
I assume this isn't what you meant.

If you mean what is 1 more than the number of  preceding siblings of the
parent, you can do

select="1+count(../preceding-sibling::*)"

David


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