How can I map a "root" Servlet so that other scripts are still runnable?

Jeremy Logan picture Jeremy Logan · Jun 23, 2009 · Viewed 31.7k times · Source

I'm trying to build a Servlet that calls a JSP page similar to the following:

public void doGet(HttpServletRequest req, HttpServletResponse resp)
        throws IOException, ServletException {
    req.getRequestDispatcher("/WEB-INF/main.jsp").forward(req, resp);
}

I need this Servlet to respond to the domain's root (eg: http://example.com/) so I'm using the following mapping in the web.xml:

<servlet-mapping>
    <servlet-name>MainServlet</servlet-name>
    <url-pattern>/*</url-pattern>
</servlet-mapping>

The problem I'm having is that this matches EVERYTHING, so when the dispatcher forwards to "/WEB-INF/main.jsp" this matches the url-pattern so the Servlet gets run again. This results in a loop that runs until it dies with a java.lang.StackOverflowError.

How can I match the root without preventing other scripts from being runnable?

Answer

nilskp picture nilskp · Feb 29, 2012

Use an empty pattern, e.g.

<servlet-mapping>
    <servlet-name>MainServlet</servlet-name>
    <url-pattern></url-pattern>
</servlet-mapping>

The servlet 3.0 spec has clarified this:

The empty string ("") is a special URL pattern that exactly maps to the application's context root

So it should at least work on a 3.0 container, and I've verified that it works on Jetty 8