Re: [xsl] correction: how to get new position() of a sorted result tree

[James Melton <james.melton@xxxxxxxxxxx>]

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() &lt; 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

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____________________________________________________________
James Melton                 CyLogix
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james.melton@xxxxxxxxxxx     www.cylogix.com

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