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From Joerg Heinicke <joerg.heini...@gmx.de>
Subject Re: xsl:copy-of Problem
Date Sun, 15 Dec 2002 11:42:35 GMT
Doesn't this sort only the first 5 elements?

<xsl:for-each select="/all/message">
   <xsl:sort select="."/>
   <xsl:if test="position() &lt; 6">
     <xsl:value-of select="."/>
   </xsl:if>
</xsl:for-each>

Joerg

Conal Tuohy wrote:
> How about something like this, then?
> 
> <xsl:variable name="m" select="/all/message"/>
> 
> <xsl:for-each select="$m[position()&lt;6">
> 	<xsl:sort select="."/>
> 	<!-- sample output here -->
> 	<xsl:value-of select="."/>
> </xsl:for-each>
> 
> Or you might want to look at:
> http://www.dpawson.co.uk/xsl/sect2/N6461.html#d6361e489
> 
> Really you should ask this kind of question on the mulberrytech xsl list;
> you'll get more response! :-)
> 
> Con
> 
>>-----Original Message-----
>>From: Marcel Jurk [mailto:marceljurk@yahoo.de]
>>Sent: Sunday, 15 December 2002 18:52
>>To: cocoon-users@xml.apache.org
>>Subject: RE: RE: xsl:copy-of Problem
>>
>>
>>Hi Con,
>>
>>thanks for the hint with the RTF. But my problem is,
>>that I want first sort the elements and afterwards I
>>want output only the first five elements.
>>With your example
>><xsl:variable name="m" select="/all/message"/>
>>is no sorting possible.
>>
>>Regards,
>>Marcel


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