Re: [xsl] Position() of parent node

Subject: Re: [xsl] Position() of parent node
From: Dimitre Novatchev <dnovatchev@xxxxxxxxx>
Date: Tue, 6 Feb 2001 10:29:45 -0800 (PST)
Hi Simon,

I guess that what you really wanted to obtain was:
count(../preceding-sibling::*) + 1

Please, note however, that the meaning of position depends of which
node-set the node is considered to be a member. A single node can have
different "positions" in different node-sets. As it was discussed in
this group just a few days ago, node-sets are considered unordered, but
there's an iherent ordering which is the "document order".

To that extent your question was not quite precise.

Dimitre Novatchev

Simon Cansick wrote:

Can anyone provide me with the syntax for getting the position() value
the current nodes' parent node (and the parent parent etc. position()
value).  I seem only able to return the current position(). 

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