I am using Tomcat 3.1.1
exampl.java program is compiled by me . The .class file is copied to
webapps/examples/servlets directory.
I launched the servlet using internet explorer5.0 ,
http://localhost:8080/examples/servlet/example
I also tried html example.html.
The message comes the page can not be displayed .
for yr. ready ref. I am giving the code below . This code is not written by
me , but copied from a book 'Developing Java Servlets' by James Goodwill. I
am in the process of learning servlets.
Please help!
Best Regards
Pramod.
> example.java
> import javax.servlet.*;
> import javax.servlet.http.*;
> import java.io.*;
> import java.util.*;
>
> public class example extends HttpServlet
> {
> public void init(ServletConfig config)throws ServletException
{
> super.init(config);
> }
>
> public void doGet(HttpServletRequest
> req,HttpServletResponse res)throws
> ServletException,IOException
> {
> doPost(req,res);
> }
>
> public void doPost(HttpServletRequest
> req,HttpServletResponse
> res)throws ServletException,IOException
> {
> res.setContentType("text/html");
> PrintWriter out=res.getWriter();
> out.println("<html>");
>
> out.println("<head><title>example</title></head>");
> out.println("<body>");
> Enumeration parameters=req.getParameterNames();
> String param=null;
> while(parameters.hasMoreElements())
> {
> param=(String)parameters.nextElement();
>
>
out.println("<bold>"+param+":"+req.getParameter(param)+
> "</bold><br>");
> }
> out.println("</body><html>");
> out.close();
> }
> public String getServletInfo()
> {
> return "example information";
> }
> }
>
> example.html
> <html>
> <head>
> <title>
> example retrieve
> </title>
> </head>
> <body>
> <form
>
action=http://localhost:8080/examples/servlet/example
> method=post>
> <table style="height:173px;width:242px">
> <tr>
> <td>last name:</td>
> <td><input name="last name" align="left"
> size="15"></td>
> </tr>
> <tr>
> <td>first name:</td>
> <td><input align="left" name="first name"
> size=15> </td>
> </tr>
> <tr>
> <td>age:</td>
> <td><input align=left name=age size=2> </td>
> </tr>
> <tr>
> <td>ssn:</td>
> <td><input align=left name=ssn size=11> </td>
> </tr>
> <tr>
> <td>dob:</td>
> <td><input align=left name=dob size=8> </td>
> </tr>
> <tr>
> <td>username:</td>
> <td><input align=left name=username
> size=8> </td>
> </tr>
> <tr>
> <td>password:</td>
> <td><input align=left name=password size=8
> type=password> </td>
> </tr>
> </table>
> <input type="submit" name="submit" value="submit">
> <input type="reset" value=reset>
> </form>
> </body>
> </html>
>
>
>
=====
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----- Original Message -----
From: Christopher K. St. John <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, December 20, 2001 10:11 PM
Subject: Re: Problem in running a servlet on tomcat
> > Purav wrote:
> >
> > I dun think so you need a super.init(config) call in
> > the new servlet API.
> >
> >
> > public void init(ServletConfig config)
> > throws ServletException
> > {
> > super.init(config);
> > }
> >
>
> The super.init(config) call is definitely required
> in this case.
>
> HttpServlet defines two "init" methods:
>
> HttpServlet.init()
> HttpServlet.init(ServletConfig config)
>
> You can override either one of them. If you
> override the one with no arguments, the HttpServlet
> saves the config argument for you. If you override
> the one that takes a ServletConfig argument, then
> you are responsible for calling super.init(config).
>
> This all sounds a lot more complicated than it
> really is, take a look at the source code if
> you're curious.
>
>
> --
> Christopher St. John [EMAIL PROTECTED]
> DistribuTopia http://www.distributopia.com
>
>
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>
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>
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LISTSERV Help: http://www.lsoft.com/manuals/user/user.html
