I've poked at Tomcat for several years now. I find myself looking for a job and 
have too much time on my hands, so I've decided to take servlets seriously and 
grasp the concepts.

Yesterday I spent a fair amount of time Googling for how to set up a <servlet> 
<servlet-mapping> pair of elements in the web.xml file on my Windows XP 
installation of Tomcat 5.5.15.

Fustrated at my inability to get these correct, I simply placed a servlet class 
I copied from a book and compiled on my machine into this directory:

C:\Apache Software Foundation\Tomcat 5.5\webapps\ROOT\WEB-INF\classes

My thought was that if I couldn't get the web.xml stuff right from looking at 
various examples and explanations I was reading, if I could get one working by 
using the full path to the servlet class, I could "reverse engineer" the 
meaning of the web.xml instructions I was reading.

The term "full path" comes from an explanation of what text ought to go into 
the <servlet-class> element child of the <servlet> element in web.xml. In 
serveral articles I found, this phrase was used, as in "the full path to the 
servlet class". To me, a "full path" means all the way back to the root 
directory of the device in question, but I get the feeling that this isn't what 
these authors meant. I rather suspect that they mean a relative path from some 
point that is the root of the Tomcat application.

So far, in the address box of my browser, I have tried:

http://localhost:8080/ROOT/WEB-INF/classes/Ch1Servlet

http://localhost:8080/WEB-INF/classes/Ch1Servlet

http://localhost:8080/classes/Ch1Servlet

http://localhost:8080/Ch1Servlet

In each case I get an HTTP Status 404 error telling me that "The requested 
resource (here you can fill in each substring above starting at the first '/') 
is not available."

Reading and re-reading has not cleared up this mystery for me. Can some tell me 
in a no-jargon way, exactly how to get this servlet to become "available"? 

Thanks.

-- 
Charles Knell
[EMAIL PROTECTED] - email

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