RE: [xsl] Change xml:lang of a skos:prefLabel [ skos, rdf, xml:lang ]

Subject: RE: [xsl] Change xml:lang of a skos:prefLabel [ skos, rdf, xml:lang ]
From: Syd Bauman <Syd_Bauman@xxxxxxxxx>
Date: Wed, 14 Oct 2009 20:39:48 -0400
>   <xsl:template match="skos:altLabel|skos:prefLabel">
>     <xsl:copy>
>       <xsl:attribute name="xml:lang">en</xsl:attribute>
>       <xsl:copy-of select="@*"/>
>       <xsl:apply-templates select="@*|node()"/>
>     </xsl:copy>
>   </xsl:template>
> 
>   <xsl:template match="*">
>     <xsl:copy>
>       <xsl:copy-of select="@*"/>
>       <xsl:apply-templates select="@*|node()"/>
>     </xsl:copy>
>   </xsl:template>
> 
> This relies on the fact that if you add the same attribute to an element
> twice, the last one wins.

1) Is that true for both XSLT 1 and 2 ? 

2) No disrespect intended, but something looks amiss here to me. I
   notice there is a separate <copy-of> for the attributes, but then
   templates are applied for the attributes, too. And what does a <copy-of>
   that's a child of <copy> do, anyway?

   Ah. To answer my own question, I just ran a tiny stylesheet with
   the above identity template on itself, and got the following result.

--------- output of this stylesheet on itself ---------
<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; version="1.0">1.0
  <xsl:template match="/">/
    <xsl:apply-templates select="node()">node()</xsl:apply-templates>
  </xsl:template>
  <xsl:template match="*">*
    <xsl:copy>
      <xsl:copy-of select="@*">@*</xsl:copy-of>
      <xsl:apply-templates select="@*|node()">@*|node()</xsl:apply-templates>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>
--------- end output ---------
   So I think that's not what M. Kay meant -- I suspect the <copy-of>
   is a holdover from a previous version or a copy-and-paste error or
   some such.

   Either that or I've missed something, in which case I'm hoping
   someone explains it!

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