Steps to create a servlet example
- Steps to create the
servlet using Tomcat server
1. Create a directory structure
2. Create a Servlet
3. Compile the Servlet
4. Create a deployment descriptor
5. Start the server and deploy the
application
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
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There
are three ways to create the servlet.
1. By
implementing the Servlet interface
2. By
inheriting the GenericServlet class
3. 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.
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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.
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DemoServlet.java
1. import javax.servlet.http.*;
2. import javax.servlet.*;
3. import java.io.*;
4. public class DemoServlet extends HttpServlet{
5. public void doGet(HttpServletRequest req,HttpServletResponse res)
6. throws ServletException,IOException
7. {
8. res.setContentType("text/html");//setting the content type
9. PrintWriter pw=res.getWriter();//get the stream to write the data
10.
11. //writing html in the stream
12. pw.println("<html><body>");
13. pw.println("Welcome to servlet");
14. pw.println("</body></html>");
15.
16. pw.close();//closing the stream
17. }}
3)Compile the servlet
For compiling the Servlet, jar
file is required to be loaded. Different Servers provide different jar files:
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Jar
file
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Server
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1)
servlet-api.jar
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Apache
Tomacat
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2)
weblogic.jar
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Weblogic
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3)
javaee.jar
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Glassfish
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4)
javaee.jar
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JBoss
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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/classesdirectory.
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.
web.xml file
1. <web-app>
2.
3. <servlet>
4. <servlet-name>aaa</servlet-name>
5. <servlet-class>DemoServlet</servlet-class>
6. </servlet>
7.
8. <servlet-mapping>
9. <servlet-name>aaa</servlet-name>
10. <url-pattern>/welcome</url-pattern>
11. </servlet-mapping>
12.
13. </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:
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<web-app> represents the whole application.
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<servlet> is sub element of <web-app> and
represents the servlet.
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<servlet-name> is sub element of <servlet>
represents the name of the servlet.
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<servlet-class> is sub element of <servlet>
represents the class of the servlet.
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<servlet-mapping> is sub element of <web-app>. It is
used to map the servlet.
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<url-pattern> is sub element of <servlet-mapping>.
This pattern is used at client side to invoke the servlet.
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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.
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After
setting the JAVA_HOME double click on the startup.bat file in apache
tomcat/bin.
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Note:
There are two types of tomcat available:
1. Apache
tomcat that needs to extract only (no need to install)
2. Apache
tomcat that needs to install
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It is
the example of apache tomcat that needs to extract only.
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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:
1. projectfolder> jar cvf myproject.war *
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:
1. http://localhost:9999/demo/welcome
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