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Re: XML to XML: how to remove stylesheet; how to add line

From: Martin Honnen <mahotrash@-----.-->
To: NULL
Date: 7/2/2004 1:35:00 PM

qluo wrote:

> How can I remove the following line from my xml output:
> 
> <?xml-stylesheet href="template.xsl" type="text/xsl"?>

That is a processing instruction that can be created with the 
<xsl:processing-instruction> element.


The following is an example of an XSLT stylesheet matching a processing 
instruction and replacing it:

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

<xsl:output method="xml" encoding="UTF-8" />

<xsl:template match="processing-instruction()[contains(., 
'href=&quot;template.xsl&quot;')]">
   <xsl:processing-instruction name="mso-application">
     <xsl:text>progid="Excel.Sheet"</xsl:text>
   </xsl:processing-instruction>
</xsl:template>

<xsl:template match="@* | node()">
   <xsl:copy>
     <xsl:apply-templates select="@* | node()" />
   </xsl:copy>
</xsl:template>

</xsl:stylesheet>

If applied to the following input

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet href="template.xsl" type="text/xsl"?>
<root>
   <?some-procinstruction god="Kibo"?>
   <whatever />
</root>

the output is

<?xml version="1.0" encoding="UTF-8"?>
<?mso-application progid="Excel.Sheet"?>
<root>
   <?some-procinstruction god="Kibo"?>
   <whatever/>
</root>

-- 

	Martin Honnen
	http://JavaScript.FAQTs.com/



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