Re: [xsl] Omit Data While Using Copy

["Trevor Majic" <majic87@xxxxxxxxxxx>]

In the end, I ended up not using Copy after all.  Although I could display 
all of the data (but still exclude the <titel> data), I ran into trouble 
trying to get the data to display in a list type format.  I was able to get 
the page to display correctly with:


<?xml version='1.0' encoding='UTF-8' ?>
<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>

<xsl:template match="kapitel">

 <h2><xsl:value-of select="titel"/></h2>

	<xsl:for-each select="hanvisning">

	<br><a><xsl:attribute name="href"><xsl:value-of

select="substring-before(substring-after(@from,'('),')')"/>.xml</xsl:attribute><xsl:value-of

select="node()"/></a></br>

 	</xsl:for-each>

  <br></br><br></br><a><xsl:attribute 
name="href">2.xml</xsl:attribute>Return to Home Page</a>

 </xsl:template>
</xsl:stylesheet>


I suspect this wasn't the best way to accomplish this, but it does work.  
Since I am just starting to use xsl, I'd prefer to learn to do things the 
correct way, rather than modifying code to make it do what I need it to.

Any suggestions to improve this code would be greatly appreciated.

Thanks again to all those who helped.

Trevor
P.S.
       Hopefully, it won't be long before I am able to offer some help to 
others!  :)

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