How to process / URL while having default processing for /js,/css,

Question

For a WebApp, I need to serve all .js, .css and all images from corresponding folders in my web app development tree ./js, ./css, ... through a default handling. So any URL like

http://www.example.com/js/x.js

should be served straight from the static files in the war.

The main servlet should deal with all requests that are not for the above. I need to be able to process requests like

http://www.example.com/PROJ/ELEM/WHATEVER

with the same unique main servlet.

So I thought I'd do this in the web.xml:

<servlet-mapping>
   <servlet-name>default</servlet-name>
   <url-pattern>/css/*,/js/*,/WEB-INF/*</url-pattern>
</servlet-mapping>

and map the main servlet like this to make some JSTL mods in a JSP file:

@WebServlet(urlPatterns="/*")
public class Main extends HttpServlet {
    protected void processRequest(HttpServletRequest request, 
                                  HttpServletResponse response)
            throws ServletException, IOException {
        request.setAttribute("test", "ok");
        request.getRequestDispatcher("/WEB-INF/index.jsp")
                            .forward(request, response);        
    }
}

When I do this I end up in a recursive loop.

Is there a way to achieve this?

Check this link :- http://stackoverflow.com/questions/870150/how-to-access-static-resources-when-using-default-servlet I think it will help you. — Rahul Jain, Jul 28 '14 at 05:39
@RahulJain Thanks for the suggestion. The accepted answer to that question didn't work for me but the next one (with 30+ votes) did. Could you make it an answer, so I can vote on it and explain further? Thanks — BxlSofty, Jul 28 '14 at 10:08

score 1 · Accepted Answer · answered Jul 28 '14 at 11:06

1

Here is the explanation with same problem. http://www.kuligowski.pl/java/rest-style-urls-and-url-mapping-for-static-content-apache-tomcat,5

answered Jul 28 '14 at 11:06

Rahul Jain

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1

This works for me. In summary: define a global filter where you detect the static paths and redirect it to the default servlet with a **chain.doFilter** then redirect all others to your smarter servlet(s) with a **forward** – BxlSofty Jul 28 '14 at 11:56

score 0 · Answer 2 · answered Jul 28 '14 at 07:07

0

This is what already happens. There is a 'default servlet' that handles any request that isn't specifically mapped to an installed servlet.

answered Jul 28 '14 at 07:07

user207421

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Thanks @EJP - but my main servlet must be mapped to "/". Hence my issues. – BxlSofty Jul 28 '14 at 07:09

score 0 · Answer 3 · answered Jul 28 '14 at 12:15

A simple variation on Rahul Jain's answer. You could do what spring MVC does for static resources : DispatcherServlet is a catch all, and it is configured to delegate to default servlet for a number or url. This may be interesting for a catch all servlet because it is often designed as a front controller that delegates actual serving to other controllers (be them servlets or not).

You can simply give the prefixes of urls that should serve static resources in a comma separated string in a context param of name static_path_prefixes and put that in your servlet :

String[] staticPathPrefixes;
RequestDispatcher defaultDispatcher;

@Override
protected void service(HttpServletRequest hsr, HttpServletResponse hsr1) throws ServletException, IOException {
    String path = hsr.getServletPath();
    for (String url: staticPathPrefixes) {
        if (path.startsWith(url)) {
            defaultDispatcher.forward(hsr, hsr1);
            return;
        }
    }
    super.service(hsr, hsr1);
}

@Override
public void init() throws ServletException {
    String urls = getServletConfig().getInitParameter("static_path_prefixes");
    staticPathPrefixes = urls.split(" *, *");
    defaultDispatcher = getServletConfig().getServletContext().
            getNamedDispatcher("default");
}

How to process / URL while having default processing for /js,/css,

3 Answers3