Path to servlet

Hi,
I created a servlet which included in a package com.abc.web.Checkin and the class is CheckinServlet.
I created a dir "abc" under /var/tomcat4/webapps/. So, /var/tomcat4/webapps/abc/index.jsp will be shown when I browsed it at http://localhost:8080/abc/index.jsp.
In the index.jsp, I have a form action=/abc/servlet/com.abc.web.Checkin.CheckinServlet
And the CheckinServlet.class was placed in /var/tomcat4/webapps/abc/WEB-INF/classes/com/abc/web/Checkin/CheckinServlet.class.
However, the browser returns that there is no such page: http://localhost:8080/abc/servlet/com.abc.web.Checkin.CheckinServlet
Why and how can I fix this problem? Where should I place the servlet and be defined in form action?
Thanks
Andrew

You would be better off creating an entry for the servlet in the web.xml for your "abc" web application. That way you could address it with a short URL. See the Tomcat examples/WEB-INF/web.xml for examples with explanation.
Download the servlet API from java.sun.com for the full details on what goes in a web.xml.
Bill

Dear Sir,
In fact, I have a web.xml file which located at /webapps/abc/WEB-INF/web.xml.
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN" "http://java.sun.com/dtd/web-app_2_3.dtd">
<web-app>
<servlet>
<servlet-name>CheckinServlet</servlet-name>
<servlet-class>com.abc.web.Checkin.CheckinServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>CheckinServlet</servlet-name>
<url-pattern>/CheckinServlet</url-pattern>
</servlet-mapping>
</web-app>
But, it still did not work. Any idea?
Regards
Andrew

In web.xml, the <servlet> element should be thought of as "back-end" and hidden.

The <servlet-mapping> on the other hand, is the opposite. It is the front-side and 'public' interface to your application classes.

So in this case, your form action tag should be submitting to "/CheckinServlet", not the fully qualified class name of what should be a private class.