Notes for computer science

For Bsc(CS)TY Sub:- JSP and Servlet

Steps to create a servlet

The steps are as follows:

Create a directory structure
Create a Servlet
Compile the Servlet
Create a deployment descriptor
Start the server and deploy the project
Access the servlet

Create a directory structures

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.

As you can see that the servlet class file must be in the classes folder. The web.xml file must be under the WEB-INF folder.

Create a Servlet

There are three ways to create the servlet.

By implementing the Servlet interface
By inheriting the GenericServlet class
By inheriting the HttpServlet class

The HttpServlet class is widely used to create the servlet because it provides methods to handle http requests such as doGet(), doPost, doHead() etc.

In this example we are going to create a servlet that extends the HttpServlet class. In this example, we are inheriting the HttpServlet class and providing the implementation of the doGet() method. Notice that get request is the default request.

firstdemo.java

import javax.servlet.http.*;
import javax.servlet.*;
import java.io.*;
public class firstdemo extends HttpServlet
{
public void doGet(HttpServletRequest req,HttpServletResponse res)

throws ServletException,IOException
{

res.setContentType("text/html");//setting the content type

PrintWriter pw=res.getWriter();//get the stream to write the data

//writing html in the stream

pw.println("<html><body>");

pw.println("Welcome to servlet");

pw.println("</body></html>");

pw.close();//closing the stream
}
}

Compile the servlet

For compiling the Servlet, jar file is required to be loaded. Different Servers provide different jar files:

Jar file	Server
1) servlet-api.jar	Apache Tomcat

Two ways to load the jar file

set classpath
paste the jar file in JRE/lib/ext folder

Put the java file in any folder. After compiling the java file, paste the class file of servlet in WEB-INF/classes directory.

Create the deployment descriptor (web.xml file)

The deployment descriptor is an xml file, from which Web Container gets the information about the servlet to be invoked.
There are many elements in the web.xml file. Here is given some necessary elements to run the simple servlet program.

web.xml file

<web-app>

<servlet-name>myservlet</servlet-name>

<servlet-class>firstdemo</servlet-class>

</servlet>

<servlet-mapping>

<servlet-name>myservlet</servlet-name>

<url-pattern>/hello</url-pattern>

</servlet-mapping>

</web-app>

Description of the elements of web.xml file

There are too many elements in the web.xml file. Here is the illustration of some elements that is used in the above web.xml file. The elements are as follows:

<web-app> represents the whole application.

<servlet> is sub element of <web-app> and represents the servlet.

<servlet-name> is sub element of <servlet> represents the name of the servlet.

<servlet-class> is sub element of <servlet> represents the class of the servlet.

<servlet-mapping> is sub element of <web-app>. It is used to map the servlet.

<url-pattern> is sub element of <servlet-mapping>. This pattern is used at client side to invoke the servlet.

Start the Server and deploy the project

To start Apache Tomcat server, double click on the startup.bat file under apache-tomcat/bin directory.

One Time Configuration for Apache Tomcat Server

You need to perform 2 tasks:

set JAVA_HOME in environment variable (It is required to start server).
Change the port number of tomcat (optional). It is required if another server is running on same port (8080).

How to deploy the servlet project

Copy the project and paste it in the webapps folder under apache tomcat.

How to access the servlet

Open browser and write http://hostname:portno/contextroot/urlpatternofservlet. For example:

http://localhost:8080/demo/hello

http://localhost ---Name of server

8080 ------ Connector port

demo ------- Context root

welcome ------- Url-pattern of servlet