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index.html [welcome file] gives HTTP Status 404 – Not Found

please see my project structure here.I am using Spring MVC here is my code.
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
when I hit run the project on server I got below error .
HTTP Status 404 – Not Found
as per the concept if we do not specify the welcome list tag then automatically index.html file will be searched.but in my case it is not working however it is working if I change the extension to .jsp.
If any one knows the answer please help me .
Note: I dont want to change the url mapping. I know there are alternatives but I want to know why this is happening and not showing me the result

SpringMVC Map Servlet to Root without Removing Content Servlet

Currently, we have "root" (/) mapped to a static index.html page, but we want to upgrade to a jsp to have dynamic content. Trying to figure out how to do this. We have content that is mapped to the default content server (e.g. /css), so we don't want to change too much.
We tried:
Changing the .html to .jsp. This resulted in a blank page.
Changing the .html to .jsp and then moving the file into the WEB-INF directory. This resulted in a 404.
Trying to subclass the DefaultServlet class that content servlet is currently mapped to. This through a 500, with a class assertion error (it checked to see if it was the same class).
Adding another servlet to that url, but it overwrote the current one.
I've searched StackOverflow, but still haven't found an answer that works.
Thanks!

If I understand your question correctly, this is trivial using Spring MVC:
<mvc:default-servlet-handler/>
And in web.xml:
<servlet>
<servlet-name>spring-mvc</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<async-supported>true</async-supported>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:/META-INF/spring/your-applicationContext-web.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring-mvc</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
Is this what you have tried already?

Just set up a controller method mapped to / that returns a view name, which is your jsp file. And make sure your view resolver is set up correctly. Any of the spring mac tutorial hellos world programs out there will show how.

Should url of Ajax's $.post match url in the web.xml

I am trying to send data from js to a servlet.
my js is in webapp\secure folder while the servlet s in java\com\ servlet folder.
I read for sending data using ajax I use
$.post("someservletname", {cityName:"hello"});
In the web.xml
<servlet>
<servlet-name>someservletname</servlet-name>
<servlet-class>java.com.someservletname</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>someservletname</servlet-name>
<url-pattern>/someservletname</url-pattern>
</servlet-mapping>
But when i put a simple sysout.print statement I do not see anything. How do i make sure that the mapping worked properly.Should i give the complete url in the post function call?

Ensure you put in the URL:
context path of your web application
Regards
Philippe

Hide .jsp extension in JSP project using servlet mapping

I am writing an application in JSP, and I need to remove the ".jsp" extension from the URL. For example, I need:
http://example.com/search.jsp?q=stackoverflow
To be:
http://example.com/search?q=stackoverflow
I know that this can be done using the ".htaccess" file, but I need some other way. I have tried the following code:
<servlet-mapping>
<servlet-name>jsp</servlet-name>
<url-pattern>*.</url-pattern>
</servlet-mapping>
However, this did not work. Does anyone have some suggestions for ways to accomplish this? Thanks in advance for any help.

With a servlet mapping you need to specify each JSP individually like follows:
<servlet>
<servlet-name>search</servlet-name>
<jsp-file>/search.jsp</jsp-file>
</servlet>
<servlet-mapping>
<servlet-name>search</servlet-name>
<url-pattern>/search</url-pattern>
</servlet-mapping>
It's easier if all those JSPs are in a common path. E.g. /app/*.
<servlet>
<servlet-name>app</servlet-name>
<servlet-class>com.example.FriendlyURLServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/app/*</url-pattern>
</servlet-mapping>
with
request.getRequestDispatcher("/WEB-INF" + request.getPathInfo() + ".jsp").forward(request, response);
This assumes the JSPs to be in /WEB-INF folder so that they cannot be requested directly. This will show /WEB-INF/search.jsp on http://example.com/app/search.
Alternatively, you can use Tuckey's URLRewriteFilter. It's much similar to Apache HTTPD's mod_rewrite.

Tomcat 6.x web.xml default and custom servlet routing

I have two servlets defined in the web.xml file, namely the default2 and myservlet. The default2 servlet is used to map the static files like the javascript and css. The myservlet is used for getting dynamic content.
<servlet>
<servlet-name>default2</servlet-name>
<servlet-class>org.apache.catalina.servlets.DefaultServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet>
<servlet-name>myservlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>
classpath:my-servlet.xml
</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
The servlet mapping is defined as follows
<servlet-mapping>
<servlet-name>myservlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>default2</servlet-name>
<url-pattern>/resources/*</url-pattern>
</servlet-mapping>
When i try to access any files under /resources, i get a 404. Any ideas why this config is not working or change this config to make it work.

Tomcat's default servlet before 6.0.30 actually serves a static resource identified by HttpServletRequest.getPathInfo(), so that /style.css will be returned when /resources/style.css is requested.
Tomcat's behavior has changed from version 6.0.30 onwards. So the original configuration from the question works in newer versions of Tomcat. See "50026: Add support for mapping the default servlet to URLs other than /. (timw)" in the changelog.
Jetty's default servlet uses a full path.

It should work fine. Are those files in real also located in the /resources folder?

Your web.xml looks correct (except I would change your <load-on-startup> constants).
Make sure that your /resources exists and is a publicly visible folder in your project path and not under /WEB-INF folder.

Try changing your url-pattern for myservlet to /, and optionally adding <mvc:default-servlet-handler /> (see here) to your Spring configuration.

Removed wrong portion of the answer as per #BalusC comment.
Set a break point in your servlet and perform a debug session. Look for the path that your servlet is picking up these files at. Make sure they match the location