RE: [xsl] example for xsl:output saxon:next-in-chain

Subject: RE: [xsl] example for xsl:output saxon:next-in-chain
From: "Michael Kay" <mike@xxxxxxxxxxxx>
Date: Thu, 10 Feb 2005 14:13:30 -0000
The best place for Saxon-specific questions is the saxon-help list or forum
at sourceforge.net.

I can't see any obvious reason why this isn't working. To help you with the
problem I would need to see the contents of all the stylesheets, and to know
how you are invoking the transformation.

Michael Kay
http://www.saxonica.com/

> -----Original Message-----
> From: uap001@xxxxxxxxx [mailto:uap001@xxxxxxxxx]
> Sent: 10 February 2005 13:10
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [xsl] example for xsl:output saxon:next-in-chain
>
> Hello,
> I am trying to use the "xsl:output
> saxon:next-in-chain" in a xslt file where I need to
> pass the output from one xsl file to the next xsl file
> to generate the final output. But, I am not able to
> make it work. It just gives me the final output which
> is same as the first xsl translation. Following is my
> xslt file :
> <?xml version="1.0" encoding="ISO-8859-1"?>
> <xsl:stylesheet
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
> xmlns:saxon="http://icl.com/saxon"; version="1.0">
>
> <xsl:import href="first.xsl"/>
> <xsl:output method="xml" indent="yes"
> saxon:next-in-chain="second.xsl"/>
> </xsl:stylesheet>
>
> With this xslt file, I only get the final output which
> is the one coming out by applying first.xml. It
> doesn't seem to be applying the second.xsl
> translation. Not sure what I am doing wrong here.
>
> What I want to do is translate my.xml to final.xml as
> follow : my.xml -> translate with first.xml ->
> firstoutput.xml -> second.xml -> finaloutput.xml
>
> I don't want the intermediate firstoutput.xml - I only
> want the finaloutput.xml.
>
> - ana
>
>
>
>
>
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