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- From: om@lgsi.co.in (Om Band)
- To: XML Developers <xml-dev@ic.ac.uk>, XML:General discussion <XML-L@listserv.heanet.ie>, Jerry <jerryk@quidnunc.com>
- Date: Tue, 30 Mar 1999 13:08:30 +0530
Hi,
Kan U Help Mi Please ?
I am developing a search engine which will have a
XML
search form linked with a Servlet. The Servlet should
take input
from the textfield of the XML form, scan the database
for matches &
generate an XML page with the results found.(Dynamically)
!
Ideally it should not make a file of that XML but
should throw
it directly to the client m/c. In this case I am
not able to link it
with the already created XSL stylesheet which will
already be
on the server (Static).
Even with making a separate XML file I am not able
to display
the XML page through Servlet, though directly it
could be displayed
with the same address typed in Address field of
the browser !!
The code I am using for Servlet is............
(This makes a separate file)
doPost(HttpServletRequest,
HttpServletResponse response)
{-------
--------
String file = "c:\\xml\\file.xml";
fw
= new FileWriter(file);
pw = new PrintWriter(fw);
pw.println("<?xml
version=.....--------"); // Making
an
pw.println("-------------------------");
// XML file.
---------------------------------------------
response.sendRedirect(file);
}
This code gives an error : xsl not found.
(Which is there in the same dir)
Alse How can I do this without making a separate
.xml file ?
THANKS !!
-Om
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