If I sort a node-set with a for-each loop, I can use:

    <xsl:value-of select="position()"/>

   to determine each node's position in the sorted list.

How do I retrieve the position of each node before it was sorted?

For example, here is a XML file I am starting with:

<?xml version="1.0" encoding="UTF-8"?>
<bar>
     <foo>D</foo>
     <foo>C</foo>
     <foo>B</foo>
     <foo>A</foo>
</bar>


I will sort the "foo's" and I want to create the following:

<?xml version="1.0" encoding="UTF-8"?>
<newRoot>
     <foo name="A" sortedPosition="1" originalPosition="4"/>
     <foo name="B" sortedPosition="2" originalPosition="3"/>
     <foo name="C" sortedPosition="3" originalPosition="2"/>
     <foo name="D" sortedPosition="4" originalPosition="1"/>
</newRoot>

How do I get the "originalPosition"?

Here is the XSL I am working with:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
version="1.0">
     <xsl:template match="/bar">
         <xsl:element name="newRoot">
             <xsl:for-each select="foo">
                 <xsl:sort order="ascending" data-type="text"/>
                 <xsl:element name="foo">
                     <xsl:attribute name="name">
                         <xsl:value-of select="."/>
                     </xsl:attribute>
                     <xsl:attribute name="sortedPosition">
                         <xsl:value-of select="position()"/>
                     </xsl:attribute>
                     <xsl:attribute name="originalPosition">
                         <xsl:value-of 
select="/some/mystery/expression()"/>
                     </xsl:attribute>
                 </xsl:element>
             </xsl:for-each>
         </xsl:element>
     </xsl:template>
</xsl:stylesheet>

Sean