What is the most straightforward way to render an XML document
differentially, using a different style sheet, dependant on the source
url

I include this directive in my xml

<?xml-stylesheet type="text/xsl"
href="..\xsl\render-specification.xsl"?>

which renders the xml through the "render-specification.xsl"  However
I want to also render some of the xml through a different xsl.

How do I accomplish this?

---

Clarification of Question by lczajka-ga on 07 Nov 2002 13:56 PST

I think so.

But I need clarification.

It looks like the following code is extracted from an xml document.
And it looks like the href parameter somehow tags the referring
request.

but how do we get to differential rendering?  Through the media
attribute?

-----------------------------------------------------------------------
<?xml version="1.0"?>
<?xml-stylesheet href="scores.xsl" type="text/xsl"?>
<?xml-stylesheet href="scores_wap.xsl" type="text/xsl" media="phone"?>
<?xml-stylesheet href="scores_print.xsl" type="text/xsl"
media="print"?>
<results showName="New Port Richey Open" location="New Port Richey,
FL" showDate="7/31/01">
...
</results>

Iczajka-ga,

http://www-106.ibm.com/developerworks/xml/library/x-tipxsltjx/listing4.html

Is that what you are looking for?

http://www-106.ibm.com/developerworks/xml/library/x-tipxsltjx/index.html

Try that link, I believe it talks about what you are asking.

-Tox-ga