> Basically from the above example, " <xsl:apply-templates 
> select="*"/> " will
> return me all the elements of the book node, however I also 
> want to retrieve
> all attributes.
> I 've tried  "   <xsl:apply-templates select="@*"/>  " to get all
> attributes, but obviously Im doing something wrong.
> Is there any simpler way of  going about sorting an xml doc 
> and returning
> the entire document in XML format (only sorted) ?

The identity transformation is...

<xsl:template match="node() | @*">
  <xsl:copy>
    <xsl:apply-templates/>
  </xsl:copy>
</xsl:template>

You could do a template for the root something like this...

<xsl:template match="/">
  <xsl:apply-templates select="node() | @*">
    <xsl:sort select="node whose value you want to use in sort"/>
  </xsl:apply-templates>
</xsl:template>

Add in the identity template and you're away in a hack.

Cheers,
Dave.

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list