Question

web.xml fragment

<!-- Handles all requests into the application -->
<servlet>
    <servlet-name>Spring MVC Dispatcher Servlet</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/app-config.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>Spring MVC Dispatcher Servlet</servlet-name>
    <url-pattern>/*</url-pattern>
</servlet-mapping>

It works fine but I’d like to NOT let the Dispatcher Servlet handle *.html requests. How do I I do that? Thanks.

Answer 1

Map it on a more specific url-pattern.

<servlet-mapping>
    <servlet-name>Spring MVC Dispatcher Servlet</servlet-name>
    <url-pattern>/spring/*</url-pattern>
</servlet-mapping>

Create a Filter which is mapped on /*.

<filter-mapping>
    <filter-name>Your Dispatcher Filter</filter-name>
    <url-pattern>/*</url-pattern>
<filter-mapping>

Which does the following in doFilter() method.

String uri = ((HttpServletRequest) request).getRequestURI();
if (uri.endsWith(".html")) {
    chain.doFilter(request, response); // Just let it go (assuming that files are in real not placed in a /spring folder!)
} else {
    request.getRequestDispatcher("/spring" + uri).forward(request, response); // Pass to Spring dispatcher servlet.
}