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ena jenkins

Subject: XSLT to sort a list alphabetically and divide it into two columns...
Author: ena jenkins
Date: 11 Mar 2009 06:39 PM
Originally Posted: 11 Mar 2009 06:40 PM

Hi,

I need to sort a list of countries alphabetically and display the list
in two columns, but I need the alphabetical list to read vertically
instead of left to right. How do I do this?

Type of output needed (Example):

Australia Finland
Belgium Hong Kong
Canada India
Chile Switzerland

Here is the XML (modified):

<countries>

<country>
<name>Hong Kong</name>
<url>/international/funding/countries/hongkong.aspx</url>
</country>

<country>
<name>Finland</name>
<url>/international/funding/countries/finland.aspx</url>
</country>

<country>
<name>Colombia</name>
<url>/international/funding/countries/colombia.aspx</url>
</country>

<country>
<name>Belgium</name>
<url>/international/funding/countries/belgium</url>
</country>

<country>
<name>China</name>
<url>/international/funding/countries/china.aspx</url>
</country>

<country>
<name>India</name>
<url>/international/funding/countries/india.aspx</url>
</country>

<country>
<name>Australia and New Zealand</name>
<url>/international/funding/countries/australia.aspx</url>
</country>

<country>
<name>Switzerland</name>
<url>/international/funding/countries/switzerland.aspx</url>
</country>

</countries>

Here is my XSLT:

<xsl:for-each select="countries/country">
<xsl:sort select="name"/>
<p>
<a>
<xsl:attribute name="href">
<xsl:value-of select="url"/>
</xsl:attribute>
<xsl:value-of select="name"/>
</a>
</p>

</xsl:for-each>

Thanks...
Any help would be greatly appreciated!

ena jenkins

Subject: XSLT to sort a list alphabetically and divide it into two columns...
Author: ena jenkins
Date: 11 Mar 2009 06:44 PM

Australia ____________ Finland
Belgium ______________ Hong Kong
Canada _______________ India
Chile ________________ Switzerland

**My spaces (tabs) used to indicate columns got stripped out when I submitted, but you get the point...

John Bampton

Subject: XSLT to sort a list alphabetically and divide it into two columns...
Author: John Bampton
Date: 12 Mar 2009 10:31 AM
Originally Posted: 12 Mar 2009 10:29 AM

Hi this should do the trick

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs"
version="2.0">
<xsl:output method="xml" indent="yes"></xsl:output>
<xsl:template match="/">
<xsl:variable name="sortedcountries">
<countries>
<xsl:for-each select="countries/country">
<xsl:sort select="name"/>
<xsl:copy-of select="."/>
</xsl:for-each>
</countries>
</xsl:variable>
<xsl:variable name="counthalf" select="count(countries/country) div 2"></xsl:variable>
<table>
<xsl:for-each select="$sortedcountries/countries/country">

<xsl:if test="position() <= $counthalf">
<tr>
<td>
<a href="{url}"><xsl:value-of select="name"/></a>
</td>
<td>
<a href="{./following-sibling::country[position() = 4]/url}"><xsl:value-of select="./following-sibling::country[position() = 4]/name"/></a>
</td>
</tr>
</xsl:if>
</xsl:for-each>
</table>
</xsl:template>
</xsl:stylesheet>

Thanks, John Bampton

James Durning

Subject: XSLT to sort a list alphabetically and divide it into two columns...
Author: James Durning
Date: 12 Mar 2009 03:12 PM
Originally Posted: 12 Mar 2009 02:21 PM

Replace the 4's with $counthalf, and change variable declaration to:
<xsl:variable name="counthalf" select="ceiling(count(countries/country) div 2)"/>

to deal with odd numbers.

There's no need to create an extra variable to hold the sorted countries; the position() function automatically updates when using the for-each and sort elements.

With these modifications, you would only need XSLT 1.0.

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