XSL numeric generate-id()

Question

How can XSL generate a unique id attribute for every element in an XML document using XSL where the id must be numeric? The XLS below works except that the generated ids are alphanumeric and I need numeric?

<?xml version='1.0' encoding='utf-8'?>  
<xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'
    xmlns:msxsl='urn:schemas-microsoft-com:xslt' exclude-result-prefixes='msxsl'>  
    <xsl:output method='xml' indent='yes'/>  
    <xsl:template match='*'>  
      <xsl:copy>  
        <xsl:attribute name='ElementID'>  
          <xsl:value-of select='generate-id()'/>  
        </xsl:attribute>  
        <xsl:apply-templates/>  
      </xsl:copy>  
    </xsl:template>    
</xsl:stylesheet>  

Thank you.

Answer 1

Switching using number() with level and count seems to have done the trick.

Thank you

<?xml version='1.0' encoding='utf-8'?>
<xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'
    xmlns:msxsl='urn:schemas-microsoft-com:xslt' exclude-result-prefixes='msxsl'>
    <xsl:output method='xml' indent='yes'/>
    <xsl:template match='*'>
      <xsl:copy>
        <xsl:attribute name='ElementID'>
          <xsl:number level='any' count='*' />
        </xsl:attribute>
        <xsl:copy-of select="@*"/><!--copy of existing all attributes-->
        <xsl:apply-templates/>
      </xsl:copy>
    </xsl:template>  
</xsl:stylesheet>

Answer 2

You can alwas use:

concat(count(ancestor::node(), '00000000', count(preceding::node())))

Knowledgeable people such as Michael Kay warn that <xsl:number/> is not efficient (sometimes O(N^2)) and should be avoided if possible.