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Creating and Running a Servlet Program

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Steps for creating and running a servlet

1. Create a directory structure
2. Create a Servlet
3. Compile the Servlet
4. Create a deployment descriptor
5. Start the server
6. Access the servlet

Create a directory structure

The directory structure defines that where to put the different types of files so that web container may get the information and respond to the client.

Create a Servlet

The servlet example can be created by three ways:
1) By implementing Servlet interface,
2) By inheriting GenericServlet class, (or)
3) By inheriting HttpServlet class

NOTE: The mostly used approach is by extending HttpServlet because it provides http request

HelloWorld.java

// Import required java libraries
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;

// Extend HttpServlet class
public class HelloWorld extends HttpServlet {

   public void doGet(HttpServletRequest request, HttpServletResponse response)
      throws ServletException, IOException {
      
      // Set response content type
      response.setContentType("text/html");

      // Actual logic goes here.
      PrintWriter out = response.getWriter();
      out.println("<h1>" + message + "</h1>");
   }
}

Note: HttpServlet contains all the life cycle methods of Servlet and the service() method is writen in the following two methods, they are
public void doGet (HttpServletRequest, HttpServletResponse) throws ServletException, IOException
public void doPost (HttpServletRequest, HttpServletResponse) throws ServletException, IOException

Compile the Servlet

To compile the servlet program we have to set a CLASSPATH

..\tomcat\lib\servlet-api.jar

Now compile the HelloWorld.java using following command

javac HelloWorld.java

If the compilation is successfull then HelloWorld.class file will be generated. Copy *.class file into document root/WEB-INF/classes folder

Create a deployment descriptor

1. Whenever client makes a request to a servlet that request is received by server and server goes to a predefined file called web.xml for the details about a servlet.
2. web.xml file always gives the details of the servlets which are available in the server.
3. If the requested servlet is available in web.xml then server will go to the servlet, executes the servlet and gives response back to client.
4. If the server is not able to find the requested servlet by the client then server generates an error (resource not found).

web.xml

<web-app>
	<servlet>
		<servlet-name>demo</servlet-name>
		<servlet-class>HelloWorld</servlet-class>
	</servlet>
	<servlet-mapping>
		<servlet-name>demo</servlet-name>
		<url-pattern>/servletdemo</url-pattern>
	</servlet-mapping>
</web-app>

Start the server

To start Apache Tomcat server, double click on the startup.bat available at ../tomcat/bin directory.

Access the servlet

Open broser and write http://hostname:portno/contextroot/url-pattern. For example:

http://localhost:8080/studyglance/servletdemo

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