Class not found exception for the servlet class.

[mazem]

I have created a servlet class with do get method in it.When am trying to run it on the tomcat server, it says class not found exception.Below is the servlet configuration in web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
<display-name>SimpleProject</display-name>
<servlet>
<description>Servlet uses Get </description>
<display-name>ServletName</display-name>
<servlet-name>ServletName</servlet-name>
<servlet-class>MyPackg.Serv.SimpleServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ServletName</servlet-name>
<url-pattern>/SimpleServletPath</url-pattern>
</servlet-mapping>
</web-app>.
When am trying to hit the URL http://localhost:8080/SimpleProject/SimpleServletPath, it says MyPackg.Serv.SimpleServlet is not found. Can some one help me with this issue.
It is able to recognize the application simpleproject but when i access simpleservletpath it throws the class not found error.

[doWhile]

It seems all your web.xml mappings name the servlet 'ServletName'...try renaming it in your web.xml, or accessing the servlet with this name.

[Tolls]

The important part for the URL used is this bit:
<url-pattern>/SimpleServletPath</url-pattern>

So the URL xxx/SimpleProject/SimpleServletPath should bring up the servlet.

Where is your servlet class file located under WEB-INF?