RE: [xsl] <xsl:call-template> error

Subject: RE: [xsl] <xsl:call-template> error
From: Jen Jiang <zhenjiang14221@xxxxxxxxx>
Date: Fri, 5 Nov 2004 10:21:45 -0800 (PST)
It's IE6.


Hi Jen,

Which browser do you use?

Cheers,
<prs/> 

-----Original Message-----
From: Jen Jiang [mailto:zhenjiang14221@xxxxxxxxx] 
Sent: Thursday, November 04, 2004 10:22 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] <xsl:call-template> error

I have an xml which is a RSS feed link:

<?xml-stylesheet href="getRSS.xsl" type="text/xsl"?>
<RSSChannels>

  <!-- RSS 0.91 feeds -->
  <RSSChannel
src="http://www.tinnitus-research.org/atr.xml"/>
  <!-- RSS 1.0 feeds -->
  <!-- RSS 2.0 feeds -->
  <RSSChannel
src="http://www.medicinenet.com/rss/specialty/Tinnitus.xml"/>
  
</RSSChannels>

I use an xslt file to transfer RSS to HTML:
<?xml version="1.0"?>
<!-- getRSS.xsl: retrieve RSS feed(s) and convert to
HTML. -->
<xsl:stylesheet
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:dc="http://purl.org/dc/elements/1.1/";
version="1.0">
	<xsl:output method="html" encoding="iso-8859-1"/>
	<xsl:template match="RSSChannels">
		<html>
			<head>
				<title>Today's Headlines</title>
			</head>
			<body>
				<xsl:apply-templates/>
			</body>
		</html>
	</xsl:template>
	<xsl:template match="RSSChannel">
		<xsl:apply-templates select="document(@src)"/>
	</xsl:template>
	<!-- Named template outputs HTML a element with href
link and RSS
       description as title to show up in mouseOver
message. -->
	<xsl:template name="a-element">
		<xsl:element name="a">
			<xsl:attribute name="href"><xsl:apply-templates
select="*[local-name()='link']"/></xsl:attribute>
			<xsl:attribute name="title"><xsl:apply-templates
select="*[local-name()='description']"/></xsl:attribute>
			<xsl:value-of select="*[local-name()='title']"/>
		</xsl:element>
	</xsl:template>
	<!-- Output RSS channel name as HTML a link inside of
h1 element. -->
	<xsl:template match="*[local-name()='channel']">
			<xsl:call-template name="a-element"/>
		<!-- Following line for RSS .091 -->
		<xsl:apply-templates
select="*[local-name()='item']"/>
	</xsl:template>
	<!-- Output RSS item as HTML a link inside of p
element. -->
	<xsl:template match="*[local-name()='item']">
			<xsl:call-template name="a-element"/>
			<xsl:text> </xsl:text>
			<xsl:if test="dc:date">
				<!-- Show date if available -->
				<xsl:text>( </xsl:text>
				<xsl:value-of select="dc:date"/>
				<xsl:text>) </xsl:text>
			</xsl:if>
	</xsl:template>
</xsl:stylesheet>

But when I ran it in browser it gave me error: 

"The XML page cannot be displayed
Cannot view XML input using XSL style sheet. Please
correct the error 
and
then click the Refresh button, or try again later. 


----------------------------------------------------------------------------
----

Keyword xsl:stylesheet may not contain
xsl:call-template."

I don't know why this happened. I have used
<xsl:call-template> in <xsl:stylesheet> so many times
(not these files) without getting the error.

Thank you very much for any input.









		
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