RE: [xsl] default template ?

Thanks Michael.
It is clearer now (I answered first to Andrew because your e-mail fall into my 
spam filter...)

Best regards,
Fabien

-----Message d'origine-----
De : Ludwig, Michael [mailto:Michael(_dot_)Ludwig(_at_)delphi-mb(_dot_)de] 
Envoyé : lundi 27 septembre 2010 17:41
À : xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Objet : RE: [xsl] default template ?

> I was wondering how an XSL Stylesheet can be done to treat 
> all nodes in an XML file if a template doesn't fit.
Two examples. The copy template:

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

A (warning) message template:

  <xsl:template match="@*|node()">
    <xsl:message>
      <xsl:value-of select="concat( '# ', name(), '&#10;' )"/>
      <xsl:apply-templates select="@*|node()"/>
    </xsl:message>
  </xsl:template>

You can think up your own default action.

Default matching is:

  <xsl:template match="@*|node()">

Michael

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