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

Posted by kalininew on Stack Overflow See other posts from Stack Overflow or by kalininew
Published on 2010-03-18T15:13:54Z Indexed on 2010/03/18 16:51 UTC
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exslt

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node-set

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position

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xpath

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xslt

Good afternoon, gentlemen. Help me solve a very simple task.

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 write code

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

and get a response error. How to do it right?

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