[xsl] Position() of parent node

[Simon Cansick <SC at access-accounts dot com>]

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
sc@access-accounts.com

Access Accounting Ltd
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