But if you don't know name of element "line"? It should be the same name
only this you know.
Thanks,
Jenya
James Carlyle writes:
Try
select="count(../preceding-sibing::line + 1)"
Kind regards,
James Carlyle
-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com]On Behalf Of
evgeniy(_dot_)strokin(_at_)rocketrainer(_dot_)com
Sent: 17 October 2002 20:11
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Re: Is it possible to know position of ancestor?
Yes, you are so right (how do you know what I meant if I didn't
know!?!? :)
But here is also the problem:
Let say we have another XML:
<root>
<some_tag/>
<line>
<a/>
<b/>
</line>
<line>
<a/>
<b/> - we are here
</line>
<root>
In your example select="count(../preceding-sibing::* + 1)" we get
3 because
it will count "some_tag" too. But we need to count only "line" elements.
How we can solve this problem?
Jenya
David Carlisle writes:
>
>> We are in tag "b",
> note that xslt works on elements (element nodes) not tags, and importat
> distinction.
>
>> We want to find out what is position of our ancestor
>> in their ancestor.
>
> I think you mean parent rather than ancestor:
>
> select="count(../preceding-sibing::* + 1)"
>
>
> David
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