I worked with the solution that Jim was kind enough to lay out
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:fo="http://www.w3.org/1999/XSL/Format"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:saxon="http://saxon.sf.net/" extension-element-prefixes="saxon">

<xsl:param name="previous-number" select="1" />


 <xsl:template match="/">
   <xsl:call-template name="example">
     <xsl:with-param name="documents"
select="collection('file:///L:/input/?select=*.xml')" />
     <xsl:with-param name="previous-number" select="$previous-number" />
   </xsl:call-template>
 </xsl:template>
<xsl:template name="example">
   <xsl:param name="documents">
   <xsl:param name="previous-number" required="yes" />

   <xsl:if test="$documents[1]">
     <xsl:variable name="current-number" as="xs:integer"
select="xs:integer($previous-number+1)" />


     <car seq="{$current-number}">
<xsl:copy-of select="@*|node()"/>
</car>


     <xsl:call-template name="example">
       <xsl:with-param name="documents" select="remove($documents, 1)" />
       <xsl:with-param name="previous-number" select="$current-number" />
     </xsl:call-template>
   </xsl:if>
</xsl:template>
</xsl:stylesheet>


And my L:/input contains 2 XML files for now. The root element of both is
<car>

On Mon, Jan 11, 2010 at 2:21 PM, David Carlisle <davidc@xxxxxxxxx> wrote:
>
>
>>  When I say $document[1], it considers all files as $documents[1].
>
> This is not possible. There is no data value in xpath that corresponds to
> multiple documents except a sequence of document nodes, and as you can
> not nest sequences, it is not possible to return a sequence of more than
> one item as a result of $documents[1].
>
> What expression did you actually use to give the impression that
> $document[1] is all files.
>
> David