Home/ Questions/Q 7707077
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-06-01T00:16:11+00:00

web.xml <?xml version=1.0 encoding=UTF-8?> <web-app id=WebApp_ID version=2.4 xmlns=http://java.sun.com/xml/ns/j2ee xmlns:xsi=http://www.w3.org/2001/XMLSchema-instance xsi:schemaLocation=http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd> <display-name>FirstProject</display-name> <servlet> <servlet-name>ServletClass1</servlet-name>

  • 0

web.xml 

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4" 
  xmlns="http://java.sun.com/xml/ns/j2ee" 
  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"  
  xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee 
                      http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
    <display-name>FirstProject</display-name>

    <servlet>
    <servlet-name>ServletClass1</servlet-name>
    <servlet-class>com.test.ServletClass1</servlet-class>
    </servlet>

    <servlet-mapping>
        <servlet-name>ServletClass1</servlet-name>
        <url-pattern>/first.*</url-pattern>
    </servlet-mapping>
    <welcome-file-list>
        <welcome-file>index.html</welcome-file>
        <welcome-file>index.htm</welcome-file>
        <welcome-file>index.jsp</welcome-file>
        <welcome-file>default.html</welcome-file>
        <welcome-file>default.htm</welcome-file>
        <welcome-file>default.jsp</welcome-file>
    </welcome-file-list>
</web-app>

first.html 

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" 
  "http://www.w3.org/TR/html4/loose.dtd">
<html>
  <head>
    <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
    <title>Test Page</title>
  </head>
  <body>
    <form name="f1" action="first" method="get">
      <input type="text" name=t/>
      <button value="Click" type="submit"></button>
    </form>
  </body>
</html>

ServletClass1.java

package com.test;
import javax.servlet.*;
import javax.servlet.http.*;
import java.io.*;

public class ServletClass1 extends HttpServlet 
{
    private static final long serialVersionUID = 1L;

    protected void doGet(HttpServletRequest request, 
      HttpServletResponse response) throws ServletException, IOException {
        response.setContentType("text/html");
        PrintWriter out= response.getWriter();  
        String msg = request.getParameter("t");
        out.println("the msg is"+msg);  
    }
}
Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    Your <form action> URL needs to match the <url-pattern> of the servlet mapping. Right now it doesn’t. Fix your servlet mapping <url-pattern> as follows:

    <url-pattern>/first</url-pattern>

    See also:

    Unrelated to the concrete problem, please make sure that you aren’t reading heavily outdated resources/books/tutorials. Your web.xml is declared conform Servlet 2.4 which is almost 9 years old already. We’re since 2.5 years already on Servlet 3.0.

    • 0