RE: [xsl] default template ?

Subject: RE: [xsl] default template ?
From: "Fabien Tillier" <f.tillier@xxxxxxxx>
Date: Mon, 27 Sep 2010 17:55:12 +0200
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.Ludwig@xxxxxxxxxxxx]
Envoyi : lundi 27 septembre 2010 17:41
@ : xsl-list@xxxxxxxxxxxxxxxxxxxxxx
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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