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Re: [xsl] question on xml-stylesheet processing instruction for browsers

2009-05-19 08:16:00
Your stylesheet is lacking an output tag which contains key meta-data
for the tranform


<?xml version="1.0" encoding="ISO-8859-1"?>

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>

<xsl:output encoding="UTF-8" indent="no" media-type="text/html" method="html" />

<xsl:template ...>
....

I am not sure what you are doing within the stylesheet..  there
certainly is no need to match the processing instruction,; that will
be handled auto-magically..  The first template need only look like so
to match the document  root:

<xsl:template match="/">
...

but I have no idea what yu or trying to achieve here, so I don't know
where it goes from there


On Tue, May 19, 2009 at 6:17 AM, Hermann Stamm-Wilbrandt
<STAMMW(_at_)de(_dot_)ibm(_dot_)com> wrote:

Hello,

is it possible to chain execution of stylesheets if opening eg.
"http://..../x.xml"; in a web browser?

By the pieces below "x.xml" will execute stylesheet "x.xsl" in the browser
and replace the processing
instruction's 'href="x.xsl" ' by 'href="y.xsl" ' in the generated output.

What appears in the browser is the output of execution of stylesheet
"x.xsl" only, no execution of "y.xsl".
If chaining stylesheet execution is possible, what is missing?


$ cat x.xml
<?xml version="1.0" encoding="ISO-8859-1"?>
<?xml-stylesheet type="text/xsl" href="x.xsl"?>
...

$ cat x.xsl
<?xml version="1.0" encoding="ISO-8859-1"?>

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>

<xsl:template match="/processing-instruction('xml-stylesheet')">
 <xsl:processing-instruction name="xml-stylesheet">
   <xsl:value-of select="substring-before(.,'href=')"/>
href="y.xsl"</xsl:processing-instruction>
</xsl:template>
...


Mit besten Grüßen / Best wishes,

Hermann Stamm-Wilbrandt
Developer, XML Compiler
WebSphere DataPower SOA Appliances
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 - michael dykman
 - mdykman(_at_)gmail(_dot_)com

 - All models are wrong.  Some models are useful.

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