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

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

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() &lt; 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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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
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