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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Sorting XML and select top 10 Play the video Subject: Sorting XML and select top 10 From: gerritjan <gerritjankoekkoek@xxxxxxx> Date: Wed, 12 Jan 2005 21:09:06 +0100 I would like to sort an XML source. Then select the first 10 of the sorted list being the 10 newest articles I've come up with this XSLT so far, but it does not work (it sorts on 'onderwerp', the subject of the for-each) <xsl:template match="onderwerpen" > <xsl:variable name="teller">0</xsl:variable> <xsl:for-each select="onderwerp"> <xsl:sort select="onderwerp/@datum" order="descending"/> <xsl:if test="$teller &lt;= 10"> <xsl:variable name="teller"><xsl:value-of select="$teller+1"/></xsl:variable> <item> <title> <xsl:value-of select="@titel"/> </title> <link> <xsl:value-of select="@menu"/> <xsl:text>_</xsl:text> <xsl:value-of select="@beveiliging"/> <xsl:text>_</xsl:text> <xsl:value-of select="@bestand"/> <xsl:text>.html</xsl:text> </link> <xsl:apply-templates select="inleidingtekst"/> <pubDate> <xsl:value-of select="@datum"/> </pubDate> </item> </xsl:if> </xsl:for-each> </xsl:template> Anybody having suggestions how to do this? Gerritjan Koekkoek Current Thread Sorting XML and select top 10 gerritjan - 12 Jan 2005 21:58:45 -0000 <= Joris Gillis - 13 Jan 2005 16:18:00 -0000 <Possible follow-ups> Ross, Douglas - 12 Jan 2005 22:48:58 -0000 <- Previous Index Next -> RE: Xalan Java 2: doesn't see, Michael Kay Thread Re: Sorting XML and select to, Joris Gillis RE: Xalan Java 2: doesn't see, Ross, Douglas Date XSL Transformation overhead, Bhaskar, Rajan Month

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