Re: [xsl] Default Rendering of HTML?

Subject: Re: [xsl] Default Rendering of HTML?
From: Shawn <sgrover@xxxxxxxxxxxxxx>
Date: Sun, 21 Nov 2004 13:42:46 -0700
I've have partial success now.

I tried two methods.  First I tried the <xsl:output> like so:

<xsl:output
	method="html"
	version="4.0"
	omit-xml-declaration="yes"
	cdata-section-elements="text"
	indent="yes"
/>

This didn't really make any difference.  Then while researching how to handle 
HTML tags in XML, I came across a reference that used xsl:copy-of, instead of 
value-of.  It occurred to me that what I was trying to do was place the 
entire node into my output, so copy-of was what I should be using.  So, I 
changed the <xsl:value-of select="text"/> line of my XSL to read <xsl:copy-of 
select="text"/> - and suddenly my HTML began to work.

I had two nodes that were to get rendered, but the first (which didn't contain 
HTML tags) didn't.  I resolved this by removing the <xsl:output> tag I had 
tried earlier.

So, I have a working solution.  But, I'm not sure if this is going to be 
suitable in all cases with HTML tags embedded in XML, or if there is a better 
way of doing this.  Any suggestions are appreciated.

Shawn

On Sunday 21 November 2004 13:13, Shawn wrote:
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
> 	<xsl:template match="/news">
> 		<xsl:apply-templates/>
> 	</xsl:template>
>
> 	<xsl:template match="item">
> 		<p>
> 			<span>
> 				<xsl:attribute name="class">newsDate</xsl:attribute>
> 				<xsl:value-of select="date"/>
> 			</span>
> 			<br/>
> 			<span>
> 				<xsl:attribute name="class">newsTitle</xsl:attribute>
> 				<xsl:value-of select="title"/>
> 			</span>
> 			<br/>
>
> 			<xsl:value-of select="text"/>
> 		</p>
> 	</xsl:template>
> </xsl:stylesheet>

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