1. Add a servlet's environment variable, I use Win2000, so as long as I am in my computer -> Properties -> Advanced -> Environment Variable, then create a ClassPath,

Because my Tomcat5.0 directory is in C: / Tomcat 5.0

So the variable value is c: / tomcat 5.0 / common / lib / servlet-api.jar

Different from Tomcat4 is Tomvat4's JAR file is servlet.jar

Restart your computer after editing environment variables

2. Write a servlet file

Import java.io.ioException;

Import java.io.printwriter;

Import javax.servlet.servletException;

Import javax.servlet.http.httpservlet;

Import javax.servlet.http.httpservletRequest;

Import javax.servlet.http.httpservletResponse;

Public class test extends httpservlet {

Protected Void Doget (HttpservletRequest Request, HttpservletResponse Response)

Throws servletexception, ioException {

PrintWriter out = response.getwriter ();

Out.println ("

this is a servlet test. ");

Out.flush ();

}

}

You can put it in any CLASSES directory, here, I choose the virtual directory D: / JSP / WebApp

Server.xml settings:

Path is a virtual directory, DOCBASE is a virtual directory path

There are many online settings of the virtual directory, and this is no longer introduced.

Compile the generation TEST.CLASS file in D: / Classes / under D: / Classes / Under

3. That is the most critical

Web.xml editing is as follows

Edit the following code in and :

test

test

test

/ test

Among them, test Description Class files are TEST

/ TEST Description The declared servlet "mapping" to address / TEST,

4. Run:

Start Tomcat, start the browser, enter http: // localhost: 8080 / WebApp / test If you see the output this is a servlet test. The written servlet is successful.

If you want to run on another directory, then the principle is the same!!