Originally posted by Siamak Saarmann: I have a servlet "http://localhost:8080/main/forwarder".
As soon as I try the URL "http://localhost:8080/main/forwarder/myinfo?i=1" I get a 404 Error on Tomcat 5.
A 404 usually means that there is a problem in the servlet configuration (web.xml). Since the first URL works, it means you have a url-pattern that matches '/forwarder' mapped to your servlet. The latter not working means that you do not have a url-pattern matching '/forwarder/myinfo' mapped in web.xml
If you want forwarder to match with any suffix, which you can then retrieve as the path info -- which what I am assuming that you wish to do given the subject line of your post -- you would map the servlet to url pattern/forwarder/*. [ July 08, 2005: Message edited by: Bear Bibeault ]