RE: [xsl] XSLT Sort - returning attributes

Subject: RE: [xsl] XSLT Sort - returning attributes
From: David Neary <David@xxxxxxxxx>
Date: Fri, 16 May 2003 13:17:51 +0200
> 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() | @*">

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"/>

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


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