Steps to create a servlet example
There are given 6 steps to create a servlet example. These steps are required for all the servers.
The servlet example can be created by three ways:
- By implementing Servlet interface,
- By inheriting GenericServlet class, (or)
- By inheriting HttpServlet class
The mostly used approach is by extending HttpServlet because it provides http request specific method such as doGet(), doPost(), doHead() etc.
Here, we are going to use apache tomcat server in this example. 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
1)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.
The Sun Microsystem defines a unique standard to be followed by all the server vendors. Let's see the directory structure that must be followed to create the servlet.
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.
2)Create a Servlet
|There are three ways to create the servlet.
The HttpServlet class is widely used to create the servlet because it provides methods to handle http requests such as doGet(), doPost, doHead() etc.
- By implementing the Servlet interface
- By inheriting the GenericServlet class
- By inheriting the HttpServlet class
|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.|
3)Compile the servlet
For compiling the Servlet, jar file is required to be loaded. Different Servers provide different jar files:
|1) servlet-api.jar||Apache Tomacat|
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.
4)Create the deployment descriptor (web.xml file)
The deployment descriptor is an xml file, from which Web Container gets the information about the servet to be invoked.
The web container uses the Parser to get the information from the web.xml file. There are many xml parsers such as SAX, DOM and Pull.
There are many elements in the web.xml file. Here is given some necessary elements to run the simple servlet program.
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.|
5)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 or JRE_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).
1) How to set JAVA_HOME in environment variable?
To start Apache Tomcat server JAVA_HOME and JRE_HOME must be set in Environment variables.
Go to My Computer properties -> Click on advanced tab then environment variables -> Click on the new tab of user variable -> Write JAVA_HOME in variable name and paste the path of jdk folder in variable value -> ok -> ok -> ok.
Go to My Computer properties:
Click on advanced system settings tab then environment variables:
Click on the new tab of user variable or system variable:
Write JAVA_HOME in variable name and paste the path of jdk folder in variable value:
There must not be semicolon (;) at the end of the path.
|After setting the JAVA_HOME double click on the startup.bat file in apache tomcat/bin.|
|Note: There are two types of tomcat available:
- Apache tomcat that needs to extract only (no need to install)
- Apache tomcat that needs to install
|It is the example of apache tomcat that needs to extract only.|
Now server is started successfully.
2) How to change port number of apache tomcat
Changing the port number is required if there is another server running on the same system with same port number.Suppose you have installed oracle, you need to change the port number of apache tomcat because both have the default port number 8080.
Open server.xml file in notepad. It is located inside the apache-tomcat/conf directory .
Change the Connector port = 8080 and replace 8080 by any four digit number instead of 8080. Let us replace it by 9999 and save this file.
5) How to deploy the servlet project
Copy the project and paste it in the webapps folder under apache tomcat.
But there are several ways to deploy the project. They are as follows:
- By copying the context(project) folder into the webapps directory
- By copying the war folder into the webapps directory
- By selecting the folder path from the server
- By selecting the war file from the server
Here, we are using the first approach.
You can also create war file, and paste it inside the webapps directory. To do so, you need to use jar tool to create the war file. Go inside the project directory (before the WEB-INF), then write:
Creating war file has an advantage that moving the project from one location to another takes less time.
6) How to access the servlet
Open broser and write http://hostname:portno/contextroot/urlpatternofservlet. For example: