public XMLResource xml(JSONPathQuery path) throws Exception {
		Object jsonValue = path.eval(this);
		return xml(jsonValue.toString());
	}
public XMLResource xml ( JSONPathQuery path ) throws Exception { Object jsonValue = path . eval ( this ) ; return xml ( jsonValue . toString ( ) ) ; }
Execute the given path query on the json and GET the returned URI expecting text/*

@param path path to the URI to follow
@return a new resource, as a result of getting it from the server in JSON format
@throws Exception