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Re: Position() of parent node
- To: SC at access-accounts dot com
- Subject: Re: [xsl] Position() of parent node
- From: Dimitre Novatchev <dnovatchev at yahoo dot com>
- Date: Tue, 6 Feb 2001 10:29:45 -0800 (PST)
- Cc: xsl-list at lists dot mulberrytech dot com
- Reply-To: xsl-list at lists dot mulberrytech dot com
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.
Cheers,
Dimitre Novatchev
Simon Cansick wrote:
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().
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