Reputation: 343
I'm trying to access servlet typing
http://localhost:8082/Libruary/controller
and Tomcat 6 gives an error 404:
HTTP Status 404 - /Libruary/controller
type Status report
message /Libruary/controller
description The requested resource (/Libruary/controller) is not available.
Apache Tomcat/6.0.26
Which means that something is wrong in web.xml file, I suppose.
So, here is my web.xml file:
<?xml version="1.0" encoding="UTF-8"?>
<web-app
version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<servlet>
<servlet-name>Controller</servlet-name>
<servlet-class>by.epam.web.libruary.jspservlet.Controller</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Controller</servlet-name>
<url-pattern>/controller</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>login.jsp</welcome-file>
</welcome-file-list>
</web-app>
I named and mapped the servlet, that's why I don't know what's incorrect. And here is project hierarchy in tomcat folder:
webapps
--Libruary
---web.xml
---...
---WEB-INF
----classes
-----by
------epam
-------web
--------libruary
---------jspservlet
----------Controller.class
I generated .class file using Eclipse. And .java files are in /scr directory, which is on the same level with classes (but it's optional, I suppose). Also I have /lib directory in this level, which contains servlet-api.jar. I don't know what to say else. Will be very grateful for your help.
Upvotes: 3
Views: 2188