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Re: [xsl] How to select a document element from the input xml file using xslt?


Subject: Re: [xsl] How to select a document element from the input xml file using xslt?
From: "M. David Peterson" <m.david@xxxxxxxxxx>
Date: Mon, 12 Jul 2004 13:10:35 -0600

Hi Alla,

Your implementation is read by the xslt processor and when it comes across <xsl:output... it doesnt know that you want it to output <xsl:output... but instead assumes you are declaring the <xsl:output../> element. As you probably already know the <xsl:output element must be a direct child of <xsl:stylesheet...> and cannot be inside of a template or anywhere after the first xsl:template element is declared... The way to accomplish what you are trying to do is to use the <xsl:element name="xsl:output"> and then add the necessary attributes using <xsl:attribute name="method">xml</xsl:attribute> etc...

Hope this helps!

Belkin, Alla wrote:
Hello,
I am using a stylesheet for modifications of the input xml file to the
different format of the xml file.
My output file should have a document element, the same as in the input
file. How to copy a document element? I can have input XMLs with
different document elements. I tried to do something like this, but it
doesn't work:

<xsl:choose>
	<xsl:when test="count(//Delete)  &gt; 0">
		<xsl:output method="xml"
doctype-system="http://delete.dtd"/>
	</xsl:when >
	<xsl:when test="count(//Delete) = 0">
		<xsl:output method="xml"
doctype-system="http://update.dtd"/>
	</xsl:when>
</xsl:choose>
Please, help

Thank you,

Alla Belkin


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