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Root "/" mades server load index.html before pass through the servlet

I have a Servlet that is mapped to the root directory "/":
<servlet>
<servlet-name>Main</servlet-name>
<servlet-class>com.motorola.triage.MainServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Main</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
In this servet there is a couple of small things that are done there, like authentication and retrieve of Google Plus information. After that, I'm doing a forward to a JSP file called "index.jsp"
req.getRequestDispatcher("index.jsp").forward(req,resp);
When I'm accessing "localhost:8080/" the static file "index.jsp" is loaded without passing through the servlet. For architecture reasons I can not change the name of index.jsp. I would like to ask if is there any way to change this behavior of the server and make it look to the servlet before look the index.jsp file.

This is occurring specifically because you used the name index.jsp.
This has been covered elsewhere, such as here and here and here.

app engine simple java webservice

I like to expose one JAVA method as a Web service that will accept POST ,strip the parameters out of it and reply accordingly. I read I have to use doPost(req,resp) , but How can I get to the servlet code? what should be in web.xml? there will not be a welcome-file ? After mapping the servlet, can I read it without the need for a index.html as start point?

create the doPost(req,resp) method in your servlet and map it to a url in web.xml
<servlet>
<servlet-name>HelloPost</servlet-name>
<servlet-class>packageName.HelloPost</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>HelloPost</servlet-name>
<url-pattern>/post-url</url-pattern>
</servlet-mapping>
then you can post your request to /post-url .You don't need to use index.html.Any url can be put in welcome file to load for the url /
To set /post-url as landing page , use
<welcome-file-list>
<welcome-file>/post-url</welcome-file>
</welcome-file-list>
you can get started here https://developers.google.com/appengine/docs/java/gettingstarted/creating

If you want to know how to set the web.xml to start the servlet then may be this will help you.

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

redirecting requests based on the body tomcat

I have a tomcat 7 application which I can get requests from external sources.
Most of them call my request like this:
http://localhost:8080/MyWeb/exRequest
and I build servlet with URL pattern inside MyWeb app.
However, one external source must send the request like this:
http://localhost:8080/
and in the body:
<xml name="test" />
Since I don't want to declare a general servlel (like tomcat default) since it means that any request will need to go through my servlet, I thought to change index.jsp of ROOT to redirect to my servlet.
Is it the best option?
Is there an option to create a default servlet that will be invoked only if there is a special parameter in the body?
EDITED
Please note that I get the requests to localhost:8080 and not localhost:8080/MyWeb - it's general to tomcat and not to a specific web app

You can't choose a servlet to invoke based on the request body, but you can set a servlet as the "welcome-file" in your web.xml.
<servlet>
<servlet-name>index</servlet-name>
<servlet-class>com.y.MyWelcomeServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>index</servlet-name>
<url-pattern>/index</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index</welcome-file>
</welcome-file-list>
If you want to preserve the "welcome" function of some existing index.jsp, your servlet could forward requests without the correct XML in the body to an index.jsp file located under the WEB-INF directory.

No, but you can create a Filter and forward/redirect to a specific servlet whenever the request meets certain conditions.
If using servlet 3.0 map it with #WebFilter, otherwise use web.xml and <filter> + <filter-mapping>. You should map it be executed before the default servlet.

Disable JSP extension processing

I have a JavaEE 1.4 web application running on WebSphere Application Server 6.0. In web.xml, there is a servlet configured to intercept all server requests:
<servlet-mapping>
<servlet-name>name</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
This works fine until I try to request something ending with *.jsp. In this case, server tries to find JSP with this name and fails with the error:
java.io.FileNotFoundException: JSPG0036E: Failed to find resource /cfvct/search_criteria.jsp
at com.ibm.ws.jsp.webcontainerext.JSPExtensionProcessor.findWrapper(JSPExtensionProcessor.java:279)
at com.ibm.ws.jsp.webcontainerext.JSPExtensionProcessor.handleRequest(JSPExtensionProcessor.java:261)
at com.ibm.ws.webcontainer.webapp.WebApp.handleRequest(WebApp.java:3226)
at com.ibm.ws.webcontainer.webapp.WebGroup.handleRequest(WebGroup.java:253)
at com.ibm.ws.webcontainer.VirtualHost.handleRequest(VirtualHost.java:229)
at com.ibm.ws.webcontainer.WebContainer.handleRequest(WebContainer.java:1970)
at com.ibm.ws.webcontainer.channel.WCChannelLink.ready(WCChannelLink.java:120)
at com.ibm.ws.http.channel.inbound.impl.HttpInboundLink.handleDiscrimination(HttpInboundLink.java:434)
at com.ibm.ws.http.channel.inbound.impl.HttpInboundLink.handleNewInformation(HttpInboundLink.java:373)
at com.ibm.ws.http.channel.inbound.impl.HttpInboundLink.ready(HttpInboundLink.java:253)
at com.ibm.ws.tcp.channel.impl.NewConnectionInitialReadCallback.sendToDiscriminaters(NewConnectionInitialReadCallback.java:207)
at com.ibm.ws.tcp.channel.impl.NewConnectionInitialReadCallback.complete(NewConnectionInitialReadCallback.java:109)
at com.ibm.ws.tcp.channel.impl.WorkQueueManager.requestComplete(WorkQueueManager.java:566)
at com.ibm.ws.tcp.channel.impl.WorkQueueManager.attemptIO(WorkQueueManager.java:619)
at com.ibm.ws.tcp.channel.impl.WorkQueueManager.workerRun(WorkQueueManager.java:952)
at com.ibm.ws.tcp.channel.impl.WorkQueueManager$Worker.run(WorkQueueManager.java:1039)
at com.ibm.ws.util.ThreadPool$Worker.run(ThreadPool.java:1475)
I need to have this request processed by the servlet, but seems server uses some JSPExtensionProcessor to process all paths ending with .jsp. Is there any way to change this behaviour?

Yes, you'll need to map your servlet to *.jsp in order to get *.jsp support redirected to your servlet.
<servlet-mapping>
<servlet-name>name</servlet-name>
<url-pattern>*.jsp</url-pattern>
</servlet-mapping>
It is normally a bad idea to have jsps accessible directly, however. Placing them in WEB-INF in some directory, then mapping an appropriate url (.do, .action, etc) to a servlet that then redirects internally to that JSP is the better practice.
So instead of typing thisUrl.jsp, the user would type thisUrl.do or thisUrl.action, and it would then get hit by the servlet to redirect to thisUrl.jsp.