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Old December 2nd, 2008, 10:25 AM
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Default Not outputting a specific node

Hi again, apologies for posting twice in a row, I will definately be sorted once this is answered.

I can include a specific node with:

<xsl:template match="nodename" >

But is it possible to actually exclude a specific node?

Or rather, if I was to start from the root:

<xsl:template match="/" >

Is there a way of actually specifying a node that I don't want to print in the output?

Thanks for your help.

Old December 2nd, 2008, 10:31 AM
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Well if you use an empty template e.g.
<xsl:template match="foo"/>
then the processor will do nothing with 'foo' elements.

  Martin Honnen
  Microsoft MVP - XML
Old December 2nd, 2008, 10:38 AM
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Ay, but the thing is I want to ignore them the first time I print the whole thing, and then print them out seperately at the bottom. Thus I already have a section for the specific node describing the format.

Old December 2nd, 2008, 10:49 AM
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Well you can then use the "mode" attribute:

<xsl:template match="foo"/>

<xsl:template match="foo" mode="bar">
  <foo>is the bar</foo>

Any instance of a normal <xsl:apply-templates/> will use the first one, but where you do want to print them out you can then do: <xsl:apply-template select="//foo" mode="bar"/>

/- Sam Judson : Wrox Technical Editor -/

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