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Re: Referencing value of non-sibling node
- From: Kim <kimba_40 at yahoo dot com>
- To: Jeni Tennison <jeni at jenitennison dot com>
- Cc: xsl-list at lists dot mulberrytech dot com
- Date: Tue, 9 Apr 2002 14:03:26 -0700 (PDT)
- Subject: Re: [xsl] Referencing value of non-sibling node
- Reply-to: xsl-list at lists dot mulberrytech dot com
Hi Jeni,
Thank you very much for your sugesstion. Yes, each <r> index is supposed to
map w/ each <mt> fr. the xml input file. I was able to make it work w/ a
similar technique (position()) but w/ diff. details.
--- Jeni Tennison <jeni@jenitennison.com> wrote:
> Hi Kim,
>
> > My specific question is when I have matched a <moid> and also
> > matched a <mt> value, how do I specify/reference the corresponding
> > indexed <r> value? In the example above, if I match <mt>mtVal4, the
> > fourth element, I would also print out the fourth <r>rVal4 value.
>
> From what I can tell, each <mt> is associated with a particular <r>
> based on its position, so the first <mt> in a <mi> is related to the
> first <r> within the <mv> of that <mi>, the second <mt> with the
> second <r> within the <mv> of the <mi> and so on. Is that accurate?
>
> If so, then in your template matching the <mt> element, you have to
> first find out the index of the <mt> within the <mi>. Now, you're
> actually applying templates to these <mt> elements in order, so this
> is quite easy -- all you have to do is store the position() of the
> current <mt> element in a variable:
>
> <xsl:variable name="index" select="position()" />
>
> and then use that index to find the <r> element within the <mt>
> element's parent <mi> element's child <mv> element with the same
> position:
>
> <xsl:value-of select="parent::mi/mv/r[$index]" />
>
> If that isn't what you were after, give us a more detailed description
> of what the "corresponding <r>" means.
>
> Cheers,
>
> Jeni
>
> ---
> Jeni Tennison
> http://www.jenitennison.com/
>
=====
Kim
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