This is the mail archive of the
xsl-list@mulberrytech.com
mailing list .
Re: XPath iteration?
- From: Jeni Tennison <jeni at jenitennison dot com>
- To: Derek Davies <ddavies at itasoftware dot com>
- Cc: XSL-List at lists dot mulberrytech dot com
- Date: Fri, 12 Jul 2002 09:25:57 +0100
- Subject: Re: [xsl] XPath iteration?
- Organization: Jeni Tennison Consulting Ltd
- References: <200207112212.g6BMCjU18704@itasoftware.com>
- Reply-to: xsl-list at lists dot mulberrytech dot com
Hi Derek,
> I'm sorry if this question is inappropriate here, but I haven't
> found a more appropriate place to ask, nor have I found a FAQ that
> addresses my question.
Don't worry; this is a good place to ask your question.
> I'm trying to come up with an XPath expression that will "iterate".
> What I have in mind is something to the effect of:
>
> /Root/SomeNode[position()!=position(/Root/SomeNode)]
>
> which would compare each SomeNode against all other SomeNode's
> except itself (there are multiple SomeNode children of /Root).
Given that all the SomeNode elements are siblings of each other, you
can collect together the other SomeNodes into a node set with:
preceding-sibling::SomeNode | following-sibling::SomeNode
I don't know what you want to do with them, though -- perhaps you
could describe what you actually want to get from the XPath?
Cheers,
Jeni
---
Jeni Tennison
http://www.jenitennison.com/
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list