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Help with XSL templates

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nomad_2k
Junior Member

United Kingdom
173 Posts

Posted - 25 May 2002 : 20:39:52

I know this forum is for asp but I haven't found an answer on any xml sites so her goes. I'm creating a content management system for my site, all the data is stored in a DB then converted to xml, merged with xsl and saved as a static html file. Now I want to use an xml tag for links like this:

<Link URL="http://somesite.com/">Some Text</Link>

the xsl template looks like this:

	<xsl:template match="Link">
		<a href="{@URL}" target="_blank">
			<xsl:value-of select="." />
		</a>
	</xsl:template>

And the question is how do I apply this template to the who xml file, so that it replaces all the <Link> tags?? I've only been able to get it to work by using <xsl:apply-template select="/Evolution/Gallery/Description/Link" /> and this replaces the first link tag and removes all the other text from arounf it.

Good things come to those who wait.
http://www.freeasphost.co.uk/evolution/

Kat
Advanced Member

United Kingdom
3065 Posts

Posted - 27 May 2002 : 05:24:29

You might want to try <xsl:copy-of select="."/> rather than xsl:value-of select="."/>.

copy-of takes the html as well as the text in between, value-of strips out the html and feed back the contents only.

KatsKorner

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mios
Junior Member

United Kingdom
101 Posts

Posted - 27 May 2002 : 09:27:25

This should do the trick, (obviously need to change the paths) but it should get you where you want.


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

<xsl:template match="/">
	<div>
		<xsl:for-each select="link">
			<xsl:apply-templates select="link"/>
		</xsl:for-each>			
	</div>	
</xsl:template>

<xsl:template match="link">
	<a>
		<xsl:attribute name="href"><xsl:value-of select="@href"/></xsl:attribute>
		<xsl:value-of select="."/>
	</a>	
</xsl:template>

</xsl:stylesheet>

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