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Re: how to get new position() of a node in a sorted result tree?
- To: xsl-list at lists dot mulberrytech dot com
- Subject: Re: [xsl] how to get new position() of a node in a sorted result tree?
- From: Wendell Piez <wapiez at mulberrytech dot com>
- Date: Wed, 01 Aug 2001 18:08:52 -0400
- Reply-To: xsl-list at lists dot mulberrytech dot com
David,
Assuming 'article' is below 'item', it looks as though you'll need to pass
the value of your position() down as a parameter (which you can do in a
template matching 'item').
There really isn't enough information here to know for sure: we need to see
a sample of input data. But position() returns the value of the node's
position in the current node set. In the template matching 'article',
that's a set of articles, in whatever order they were selected in. Since we
don't see articles being selected here (and we can't assume defaults
without seeing the structure of the input or your assurance that there are
no other templates selecting it), it appears that this is, naturally,
document order.
Please provide more information and I'm sure someone can help.
Good luck,
Wendell
At 05:33 PM 8/1/01, you wrote:
>Hi,
> Anyone knows how to get the new node position() of a sorted result tree?
> Specifically, I have the following XSLT code:
> <xsl:template match="/">
> <xsl:apply-templates select="//item">
> <xsl:sort data-type="number" order="descending"
> select="@date" />
> </xsl:apply-templates>
> </font>
> </xsl:template>
>
> <xsl:template match="article">
> <!-- process only the first 100 -->
> <xsl:if test="position() < 100">
> ......
> </xsl:if>
> </xsl:template>
>
> When I call the position() function, it returns the position ID in
> the original
>tree not the position ID in the new sorted tree. How to get the new
>position in a
>sorted tree?
> Thanks,
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Mulberry Technologies, Inc. http://www.mulberrytech.com
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Suite 207 Phone: 301/315-9631
Rockville, MD 20850 Fax: 301/315-8285
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