Hi there, I am running a servlet using Tomcat 4.0.1 on Windows XP Professional. The servlet is called Conti and resides in the following directory structure on my home pc: C:\jakarta-tomcat-4.0.1\webapps\conti\WEB-INF\classes I have created a web.xml file which does have the correct servlet name in the servlet tags as shown here: <?xml version="1.0" encoding="ISO-8859-1"?> <!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>Conti</servlet-name> <servlet-class>Conti</servlet-class> </servlet> </web-app> The servlet compiles with no errors. When i enter the following url in my address bar: http://localhost:8080/conti/servlet/Conti I recieve the following error: Apache Tomcat/4.0.1 - HTTP Status 404 - /conti/servlet/Conti type Status report message /conti/servlet/Conti description The requested resource (/conti/servlet/Conti) is not available.
I have recompiled my servlet and restarted Tomcat several times already. Could someone please help me. Thanks Fathima Khan.
Have you done servlet mapping in you deployment descriptor(web.xml)? if not do it as following <servlet-mapping> <servlet-name>Conti</servlet-name> <url-pattern>/conti</url-pattern> </servlet-mapping> and now try to run the servlet using http://localhost:8080/conti may be this will help.