A friendly place for programming greenhorns!
Big Moose Saloon
Register / Login
Win a copy of
Java 8 in Action
this week in the
Simple servlet implementation
Joined: Feb 24, 2006
Feb 24, 2006 06:35:00
I am a beginner in the
world. I have installed Tomcat4.1 in my computer. Can any one tell me how to implement the "Helloworld" servlet with a link to a simple html "submit" button? It would be a great help for a beginner.
Joined: Jul 26, 2005
Feb 24, 2006 08:14:00
I would recommend you to read the Head First
In short though, you need to set up TOMCAT_HOME, JAVA_HOME in your environment variables. Google for more info on those variables.
You need to have a Deployment Descriptor named web.xml which provides the mapping between your form(HTML page) and the Servlet.
more info :
Joined: May 09, 2002
Feb 24, 2006 10:21:00
1) in the webapps folder make "mysite" folder, in that make "WEB-INF" folder and in that classes forlder
2) create a simple html file with action set to "http://localhost:8080/mysite/servlet/HelloServlet" and save it as index.html
the code will be...
<form action="http://localhost:8080/mysite/servlet/HelloServlet" method=get>
<input type=submit value="Go To Servlet">
3) create and compile a simple servlet
This servlet must have a doGet() method which print out the "Hello World"
message to client. This servlet class has to be in classes folder.
4) have web.xml (web descriptor) file in WEB-INF folder
5) go to your html page by typing
in the web browser after starting
For details do read the recommended book...
Help gets you when you need it!
I agree. Here's the link:
subject: Simple servlet implementation
how to build a ajax response
All times are in JavaRanch time: GMT-6 in summer, GMT-7 in winter
| Powered by
Copyright © 1998-2014