Sorted exsl:node-set. Return node by it position

Question

I have a set of nodes

<menuList>
  <mode name="aasdf"/>
  <mode name="vfssdd"/>
  <mode name="aswer"/>
  <mode name="ddffe"/>
  <mode name="ffrthjhj"/>
  <mode name="dfdf"/>
  <mode name="vbdg"/>
  <mode name="wewer"/>
  <mode name="mkiiu"/>
  <mode name="yhtyh"/>
  and so on...
</menuList>

I have it sorted now this way

 <xsl:variable name="rtf">
    <xsl:for-each select="//menuList/mode">
       <xsl:sort data-type="text" order="ascending" select="@name"/>
          <xsl:value-of select="@name"/>
    </xsl:for-each>
 </xsl:variable>

Now I need to get an arbitrary element in the sorted array to the number of its position. I'm using the code:

<xsl:value-of select="exsl:node-set($rtf)[position() = 3]"/>

and I get a response error. How should I be doing it?

Answer 1

There are at least two errors in the provided code:

1. <xsl:value-of select="@name"/>

When more than one adjacent text node exist, they are combined into one. The result is that the RTF has just one (long) single text node, and there isn't a 3rd node.

2.<xsl:value-of select="exsl:node-set($rtf)[position() = 3]"/>

This asks for the third node contained in exsl:node-set($rtf), however exsl:node-set($rtf) is the document node of the temporary tree produced by the exsl:node-set() extension function -- this is only one node. Therefore the above XPath expression doesn't select anything at all.

One correct solution is the following:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:msxsl="urn:schemas-microsoft-com:xslt"
 >

 <xsl:template match="/">
    <xsl:variable name="rtf">
        <xsl:for-each select="//menuList/mode">
           <xsl:sort data-type="text" order="ascending" select="@name"/>
              <xsl:copy-of select="."/>
        </xsl:for-each>
     </xsl:variable>
    <xsl:value-of select="msxsl:node-set($rtf)/*[position()=3]/@name"/>
 </xsl:template>
</xsl:stylesheet>