code: public XMLResource xml(XPathQuery path, Content aContent) throws Exception {
  		String uri = path.eval(this, String.class);
  		return xml(uri, aContent);
  	}
docs: Execute the given path query on the XML, POST the returned URI expecting XML
  
  @param path path to the URI to follow, must be a String QName result
  @param aContent the content to POST
  @return a new resource, as a result of getting it from the server in text format
  @throws Exception