run a servlet program in tomcat 6.0

Question

How i can run a servlet program in tomcat 6.0?

Answer 1

By building a webapp and putting it into the webapp root, just as you would in another version of tomcat or any other servlet container.

Answer 2

First, you need to declare your servlet in a web deployment descriptor (a web.xml file) which looks like this:

<?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_2_5.xsd"
  version="2.5">
  <display-name>Archetype Created Web Application</display-name>
  <servlet>
    <servlet-name>HelloWorldExample</servlet-name>
    <servlet-class>cnx.mywebapp.HelloWorldExample</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>HelloWorldExample</servlet-name>
    <url-pattern>/hello</url-pattern>
  </servlet-mapping>
</web-app>

Basically, the idea is to declare the fully qualified name of your servlet in the servlet element and to map it to an url pattern in the servlet-mapping (the mapping is done via a unique servlet name)

Then, you need to package the whole (the servlet .class file and the deployment descriptor) in a web application archive (with a .war extension) which has a defined structure:

mywebapp
|-- WEB-INF
|   |-- classes (java classes, including your servlet, go here)
|   |-- lib     (jar dependencies go here)
|   `-- web.xml (this is the deployment descriptor) 
`-- index.jsp

Finally, deploy (copy) the .war in the webapps directory of Tomcat. To access the servlet:

http://localhost:8080/mywebapp/hello
           A       B     C       D

Where:

• A is the hostname where Tomcat is running (the local machine here)
• B is the port Tomcat is listening to (8080 is the default)
• C is the context path to access the webapp (the name of the war by default)
• D is the pattern declared in the web.xml to invoke the servlet