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Re: correction: how to get new position() of a sorted result tree
- To: xsl-list at lists dot mulberrytech dot com
- Subject: Re: [xsl] correction: how to get new position() of a sorted result tree
- From: James Melton <james dot melton at cylogix dot com>
- Date: Thu, 02 Aug 2001 11:35:25 -0400
- References: <3B69671F.B51F74E7@mediaone.net>
- Reply-To: xsl-list at lists dot mulberrytech dot com
There's a little bug in your sort select. You want select="date" instead
of select="@date"
Jim.
David Li wrote:
>
> Hi,
> I apologize for the typo in my previous question. The original XSLT code
> is:
> <xsl:template match="/">
> <xsl:apply-templates select="//article">
> <xsl:sort data-type="number" order="descending" select="@date" />
> </xsl:apply-templates>
> </xsl:template>
>
> <xsl:template match="article">
> <xsl:if test="position() < 100">
> ...... <!-- do processing here -->
> </xsl:if>
> </xsl:template>
>
> The original XML source file is:
>
> <?xml version="1.0" encoding="ISO-8859-1"?>
> <index section="Headlines">
> <article filename="file1.xml">
> <title>News 1 Title Text</title>
> <date>20001020</date>
> <ctprovider>Efe</ctprovider>
> </article>
> <article filename="file2.xml">
> <title>News 2 Title Text</title>
> <date>20001113</date>
> <ctprovider>Eastside Journal</ctprovider>
> </article>
> <article filename="file3.xml">
> <title>News 3 Title Text</title>
> <date>20001113</date>
> <ctprovider>Newsbytes News Network</ctprovider>
> </article>
> ............ <!-- more article elements here -->
> </index>
>
> The question is how to get the position() from a sorted result tree. In
> the above code, calling position() returns the ID of the original pre-sorted
> tree.
> Thanks,
>
> David
>
> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
--
____________________________________________________________
James Melton CyLogix
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james.melton@cylogix.com www.cylogix.com
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