Re: xsl:copy-of Problem

[Joerg Heinicke]

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:[EMAIL PROTECTED]]
Sent: Sunday, 15 December 2002 18:52
To: [EMAIL PROTECTED]
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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