Spring Websocket, 404 error while connecting. App not using Spring MVC - spring

Question

I'm trying to get working the Spring Example tutorial:
https://spring.io/guides/gs/messaging-stomp-websocket/
But I need it working on an existing Application that I need to add WebSocket support. The idea is once I get the basic working, start building from there.
The differences I have with the example in the URL is:
I'm not using SpringBootApplication, but Tomcat instead (7.0.69)
That implies I do have a web.xml (included below)
I skipped the Application.java with the main... I'm building a WAR and deploying it manually into Tomcat.
Whem I start Tomcat I read the following which seems relevant:
03:06:52,908 INFO SimpleBrokerMessageHandler:157 - Starting...
03:06:52,908 INFO SimpleBrokerMessageHandler:260 - BrokerAvailabilityEvent[available=true, SimpleBrokerMessageHandler [DefaultSubscriptionRegistry[cache[0 destination(s)], registry[0 sessions]]]]
03:06:52,913 INFO SimpleBrokerMessageHandler:166 - Started.
The problem is the following:
When I click on connect on the index.html, I get a 404 when I'm trying to reach the server at the mapping url "/hello" (not sure why info is there, I guess is part of the protocol..) http://localhost:8080/hello/info
I was googling this, even found some answers in stackoverflow. Some have fixed this addding as prefix the DispatcherServlet mapping to the websocket url...
However my app isn't using Spring MVC, so I don't have a DispatcherServlet configured... and I looks like an overkill to include it just for the WebSockets.
I tried adding the annotation #EnableWebMvc to the WebSocketConfig class (in the link is the code, I have the same lines)
Any suggestion will be much appreciated !
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_3_0.xsd"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
id="WebApp_ID" version="3.0">
<display-name>App</display-name>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/classes/applicationContext.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<listener>
<listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
</listener>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
</web-app>

Answer 1

I give up trying to avoid having an MVC Dispatcher for an APP witch doesn't use Spring MVC.
This solved it:
web.xml
<servlet>
<!-- Only used by websockets -->
<servlet-name>SpringMVCDispatcherServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/classes/websockets-application-config.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>SpringMVCDispatcherServlet</servlet-name>
<url-pattern>/ws/*</url-pattern>
</servlet-mapping>
And then, in the index.html I only added the prefix /ws/ when connecting (only there, sinding messages without the prefix is ok).