title: Tomcat: correct way to find a resource?
body: <p>I've working on a big application deployed on a Tomcat server.  There's only one .jsp page and all the UI is done using ext-js.</p>
  
  <p>There are a lot of .java classes.  In one of these class (which is performing validation), I'd like to add XML validation and to do this I need to access the <em>.xsd</em> file.</p>
  
  <p>My problem is that I don't know how to cleanly find the path to my .xsd file.</p>
  
  <p>I'll put the .xsd files in a repertory next to css/, images/, etc.</p>
  
  <p>I know how to this in a crappy way: I call <em>System.getProperty("user.home")</em> and from there I find my webapp, the <em>xsd/</em> folder, and the <em>.xsd</em> files.</p>
  
  <p>But what is a "clean" way to do this?</p>
  
  <p>Where am I supposed to find the path to my webapp (or to my webapp resources) and how am I supposed to pass this information down to the <em>.java</em> class that performs the validation?</p>