Home/ Questions/Q 7187235
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-28T18:48:00+00:00

this is my first servlet ever. here is it’s code. import javax.servlet.*; import javax.servlet.http.*;

  • 0

this is my first servlet ever. here is it’s code.

import javax.servlet.*;
import javax.servlet.http.*;
import java.io.*;
public class Ch1Servlet extends HttpServlet {
public void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException {
    PrintWriter out = response.getWriter();
    java.util.Date today = new java.util.Date();
    out.println("<html> " +"<body>" +"<h1 align=center>HF\'s Chapter1 Servlet</h1>" +" " + "<br>" + today + "</body>" + "</html>");
    }
}

I compiled it using this command
javac -classpath /usr/share/tomcat7/common/lib/servlet-api.jar -d classes src/Ch1servlet.java
I then put the .class file in the classes folder in my WEB-INF folder.

Here is my web.xml

<?xml version="1.0" encoding="ISO-8859-1" ?>
<web-app 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" version="2.4">
    <servlet>
        <servlet-name>Chapter1 Servlet</servlet-name>
        <servlet-class>Ch1Servlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>Chapter1 Servlet</servlet-name>
        <url-pattern>/Serv1</url-pattern>
    </servlet-mapping>
</web-app>

Tomcat7 keeps giving me a 404 on http://127.0.0.1:8080/ch1/Serv1/ saying The requested resource (/ch1/Serv1/) is not available.

File Tree:
File Tree

What am i doing wrong here?

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    You should put servlet classes in a package. Whether packageless servlets works depend on the specific combination of an older Tomcat and JVM version. If you see this example in a book/tutorial, then it is surely far outdated.

    package com.example;

// ...

public class Ch1Servlet extends HttpServlet {
    // ...
}

    You should have a /com/example/Ch1Servlet.java file. Compile it as follows

    javac -classpath /usr/share/tomcat7/common/lib/servlet-api.jar -d classes src/com/example/Ch1servlet.java

    (I however wonder what the common lib is doing there, this was typical for Tomcat 4.x/5.x, but it’s not present since Tomcat 6. If you manually changed Tomcat’s structure in order to follow the instructions of an outdated tutorial, undo it!)

    Put the com folder with the generated class by its entirity in /WEB-INF/classes folder of your webapp. So you must have a /WEB-INF/classes/com/example/Ch1Servlet.class.

    Then, edit your /WEB-INF/web.xml to specify the fully qualified name (FQN) of the servlet class in <servlet-class>:

    <?xml version="1.0" encoding="UTF-8"?>
<web-app 
    xmlns="http://java.sun.com/xml/ns/javaee" 
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
    version="3.0" 
>
    <servlet>
        <servlet-name>Chapter1 Servlet</servlet-name>
        <servlet-class>com.example.Ch1Servlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>Chapter1 Servlet</servlet-name>
        <url-pattern>/Serv1</url-pattern>
    </servlet-mapping>
</web-app>

    (please note that I fixed the root declaration as well to comply Tomcat 7 supported servlet version, it would otherwise fall back to least compatibility modus)

    • 0