[xsl] Position() of parent node

Subject: [xsl] Position() of parent node
From: Simon Cansick <SC@xxxxxxxxxxxxxxxxxxx>
Date: Tue, 6 Feb 2001 18:09:33 -0000
Can anyone provide me with the syntax for getting the position() value of
the current nodes' parent node (and the parent parent etc. position()
value).  I seem only able to return the current position(). 

I need to use the value in my generated HTML, not for template selection.

Thanks for any help.

Simon Cansick

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