Hi,
I asume that You are using URLConnection to the
Servlet.
When i have used application -servlet comunication
i have just sent with the PrintWriter the word "ok".
PrintWriter toClient=res.getWriter();
toClient.println("ok");
toClient.close();
Now in my applet(or application) i read the inputStream:
BufferedReader in = new BufferedReader(new
InputStreamReader(conn.getInputStream()));
String inputLine;
while ((inputLine = in.readLine()) != null)
{
if(inputLine.indexOf("ok") >=0 ) success=true;
break;
}
Now in Your applet You can deside that everything worked fine at server side
(meaning the
Servlet has done his job).
You can open a html page from the applet. (sorry i don't know the exact
method for this).
Maybe somebody else??
Hope it helps.
Best wishes,
Andras
----- Original Message -----
From: rajeshcm <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, November 18, 1999 7:13 AM
Subject: Servlet Problem
> > Servlet Response Problem
> >
> >
> > I am using a Servlet to process a request from Applet. I am sending the
> > parameters from the applet to Servlet. These are getting passed to the
> > Servlet and am able to print out all the parameters in the servlet. Now
my
> > requirement is to show back a HTML page indicating that these parameters
> > were passed successfully.
> >
> > I have tried the following methods both unsuccessfully:
> > 1) using the response.sendRedirect("http://........./Status.html");
> > 2) using the response.setContentType("text/html");
> > PrintWriter pw = response.getWriter();
> > pw.println("<html>");
> > ........ all the other html code is written here
> > pw.println("parameters received OK");
> > ........ all the other html code is written here
> >
> > pw.close():
> >
> > Can someone please help me solve this problem
> >
> > thanks
> >
> > Sai Prasad
> > Pyxis Technology Solutions Pvt Ltd
> >
> > email : [EMAIL PROTECTED]
> >
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