Its really quite easy Thierry,
In the ../tomcat/conf/server.xml file there are "context"s that equate to
what IIS calls a "web". These handle the mappings between the physical disk
location and the browser url (address box) location, e.g.
"http://computer:8080/upload"; == "C:\tomcat\webapps\upload".
Then, within any one of these contexts there exists a "WEB-INF/classes"
directory. Classes placed in that directory are typically (this is a high
level view) servlets and accessible simply by adding "/servlet" to the
context url.
So, the physical path to:
"http://computer:8080/upload/servlet/UploadServlet";
would probably be:
"C:\tomcat\webapps\upload\WEB-INF\classes\UploadServlet.class"
-Richard
On Tuesday 31 July 2001 05:24 am, you wrote:
> Hi,
>
> First of all, sorry if my question is not clear, coz i'm not a developper
> (i don't understand a thing in java, servlet,..) but i have to manage a
> computer running IIS 4.0 + Tomcat 3.2.3 and i don't know what to do.
>
> One of our programmer created a little html form page wich does this :
>
> <form action="http://computer:8080/upload/servlet/UploadServlet";
> method="post" enctype="multipart/form-data">
>
> ok
> i have a servlet called UploadServlet.class somewhere.
> How do Tomcat know how to resolve the path in order to load the servlet ?
> I read that there are mapping in the server.xml, there are context in the
> uriworkermap.properties,...
>
> I'm completly lost. I want to understand the different steps from the call
> of the Tomcat engine, to the execution of the servlet (wich actions are
> perform according to what parameters, environnement variable, registry,...)
>
> I really need your help, please.
> Thank you for your patience.
--
Richard Draucker [EMAIL PROTECTED]
Protected-Data.Com www.protected-data.com
Remote Data Support For Web Developers
