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Re: [xsl] Omit Data While Using Copy

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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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