Query on sorting XML using XSLT and getting the same XML as output !

Question

Hi,
Looking for one information regarding sorting XML using XSLT , with the sorted XML as output. For eg. my XML is :
<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="sort1.xsl"?>
<levelone>
     <child ID="1" sort="5">
          <name>Paul</name>
     </child>
     <child ID="2" sort="1">
          <name>Adam</name>
     </child>
     <child ID="3" sort="2">
          <name>Will</name>
     </child>
</levelone>
and XSL :
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/levelone">
     <xsl:copy>
     <xsl:apply-templates>
          <xsl:sort select="@sort"/>
     </xsl:apply-templates>
          </xsl:copy>
     </xsl:template>
     <xsl:template match="child">
          <xsl:copy-of select="."/>
     </xsl:template>
</xsl:stylesheet>
This does the sort based on Name. But I want to get the same xml as output with the name sorted. Eg.
<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="sort1.xsl"?>
<levelone>
     <child ID="2" sort="1">
          <name>Adam</name>
     </child>
     <child ID="1" sort="5">
          <name>Paul</name>
     </child>
     <child ID="3" sort="2">
          <name>Will</name>
     </child>
</levelone>
Any pointers will be highly appreciated.
- Thanks

Answer 1

Don't you want <xsl:sort select="name"/> rather than <xsl:sort select="@sort"/>?