-1

Actually this is my first servlet application.just getting started..
below is my code:

package newpackage.org;    

import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;

import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

//@WebServlet(description = "a general servlet", urlPatterns = { 
"/simpleservlet" })

public class simpleservlet extends HttpServlet {

    private static final long serialVersionUID = 1L;

    /**
     * @see HttpServletdoGet(HttpServletRequest request, HttpServletResponse 
response)
     */

    protected void doget(HttpServletRequest request, HttpServletResponse 
response) throws ServletException, IOException {

        // TODO Auto-generated method stub

    //  response.getWriter().append("Served at: ").append(request.getContextPath());

     PrintWriter obj= response.getWriter();;

//   String username= request.getParameter("username");

     obj.println("again getting back with  ");
    }
    }`

XML

<?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" 
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee 

http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">

  <display-name>myproject</display-name>

  <servlet-mapping>
<servlet-name>xmlservlet</servlet-name>
<url-pattern>/002path</url-pattern>
</servlet-mapping>
<servlet>

<servlet-name>xmlservlet</servlet-name>
<servlet-class>newpackage.org.xmlservlet </servlet-class>
</servlet>

<servlet-mapping>
<servlet-name>simpleservlet</servlet-name>
<url-pattern>/001path</url-pattern>
</servlet-mapping>
<servlet>

<servlet-name>simpleservlet</servlet-name>
<servlet-class>newpackage.org.xmlservlet </servlet-class>
</servlet>

</web-app>

ERROR:

HTTP Status 404 - /myproject/servlet/newpackage.org.simpleservlet

type Status report

message /myproject/servlet/newpackage.org.simpleservlet

description The requested resource is not available.

Apache Tomcat/7.0.72

mmcglynn
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Bhuban
  • 1

2 Answers2

0

You need to remove the below in web.xml

<servlet-mapping>
 <servlet-name>simpleservlet</servlet-name>
 <url-pattern>/001path</url-pattern>
</servlet-mapping>
<servlet>
 <servlet-name>simpleservlet</servlet-name>
 <servlet-class>newpackage.org.xmlservlet </servlet-class>
</servlet>

And use the below code in servlet:

@WebServlet(description = "a general servlet", urlPatterns = {"/simpleservlet", "/001path" })
Rohit Gaikwad
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0
<servlet-name>simpleservlet</servlet-name>
<servlet-class>newpackage.org.xmlservlet </servlet-class>
</servlet>

Give simpleservlet class name instead of xmlservlet

<servlet-name>simpleservlet</servlet-name>
<servlet-class>newpackage.org.simpleservlet </servlet-class>
</servlet>
Ramya
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