Java JSP/Servlet: controller servlet throwing the famous stack overflow
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Published on 2010-06-01T02:32:46Z
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2010/06/01
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I've read several docs and I don't get it: I know I'm doing something wrong but I don't understand what. I've got a website that is entirely dynamically generated: there's hardly any static content at all.
So, trying to understand JSP/Servlet, I've written my own "front controller" intercepting every single query, it looks like this:
<servlet-mapping>
<servlet-name>defaultservlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
Basically I want any user request, like:
- example.org
- example.org/bar
- example.org/foo.html
to all go through a default servlet which I've written.
The servlet then examines the URI and find to which .jsp the request must be dispatched, and then does, after having set all the attributes correctly, a:
RequestDispatcher dispatcher = getServletContext().getRequestDispatcher("/WEB-INF/jsp/index.jsp");
dispatcher.forward(req, resp);
When I'm using a url-pattern (in web.xml) like, say, *.html
, everything works fine. But when I change it to /*
(to really intercept everything), I enter an endless loop and it ends up with a... StackOverflow :)
When the request is dispatched, is the URI ".../WEB-INF/jsp/index.jsp" itself matched by the web.xml filter /* that I set?
How should I do if I want to intercept everything using a /* url-pattern and yet be able to dispatch/forward/?
I'm not asking about specs/Javadocs here: I'm really confused about the bigger picture and I'd need some explanation as to what could be going on.
Am I not supposed to intercept really everything?
If I can intercept everything, what should I be aware of regarding forwarding/dispatching?
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