Call index.html in Tomcat - spring

Question

I'm trying to run an application in Tomcat. I create an application that generates a war file that I put on the Tomcat to run the application but when I try to run
http://localhost:8080/AppletTest/
it gives me error:
HTTP Status 404 -
type Status report
message
description The requested resource is not available.
Apache Tomcat/8.0.28
And more detailed:
06-Nov-2015 16:35:07.052 WARNING [http-nio-8080-exec-51] org.springframework.web.servlet.PageNotFound.noHandlerFound No mapping found for HTTP request with URI [/AppletTest/] in DispatcherServlet with name 'mvc-dispatcher'
In the dir installation of tomcat my app resides on
~/apache-tomcat-8.0.28/webapps/
and the index.html and the other files are on the
~/apache-tomcat-8.0.28/webapps/AppletTest/WEB-INF/pages/index.html
.
I'm making any error?
My files:
web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app
version="3.0"
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
<display-name>Test</display-name>
<servlet>
<servlet-name>spring-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/mvc-dispatcher-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<load-on-startup></load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/mvc-dispatcher-servlet.xml</param-value>
</context-param>
</web-app>
mvc-dispatcher-servlet.xml:
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd">
<context:component-scan base-package="com.something.controller" />
<bean
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix">
<value>/WEB-INF/pages/</value>
</property>
<property name="suffix">
<value>.html</value>
</property>
</bean>
</beans>

Answer 1

You have created a Spring application.
On your Web.xml you have mapped all requests (/*) to your Spring Servlet.
Now, you have to create a controller that will listen to your URL.
Example:
#Controller
public class IndexController {
#RequestMapping(value = "", method=RequestMethod.GET)
public String index(Model m) {
return "index/index";
}
}
Now this method will listen the {context}/ url and will return the index.html view.