CODEWITHSUNDEEP
javaservletstomcat7
Abhishekkumar
Abhishekkumar

Reputation: 1102

Servlet program showing HTTP STATUS 404

I have written this code for servlet 

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;


public class Httpservlet1  extends HttpServlet 
{
    public void doGet(HttpServletRequest request,
    HttpServletResponse response)
    throws ServletException, IOException
    {
        String color = request.getParameter("color");
        response.setContentType("text/html");
        PrintWriter pw = response.getWriter();
        pw.println("<B>The selected color is: ");
        pw.println(color);
        pw.close();
    }
}

i have compiled it and its corresponding html file action attribute value is

action="http://localhost:8765/HS/HTTPSERVLET">

and web.xml contains

servlet-name four
servlet-class Httpservlet1

servlet-name four
url-pattern /HTTPSERVLET in xml code format still its showing error message on running it

enter image description here

Upvotes: 1

Views: 488

Answers (1)

PbxMan
PbxMan

Reputation: 7625

Perhaps your web.xml is not properly formed. It works OK for me.

    <?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
   <servlet>
    <servlet-name>four</servlet-name>
    <servlet-class>Httpservlet1</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>four</servlet-name>
    <url-pattern>/HTTPSERVLET</url-pattern>
  </servlet-mapping>
</web-app>

Upvotes: 2

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