AJAX的URL咨询(前端EXTJS)

有没有什么办法能够让EXTJS中ajax传值时的url指向后台java类中的具体方法,以前都是用了框架直接指向strus2,现在项目没用框架,求详解!

public class HelloWorld extends HttpServlet {
public void service(HttpServletRequest request,HttpServletResponse response) throws ServletException,IOException {
response.setContentType("text/html");//告诉服务器是html
PrintWriter out = response.getWriter();
out.println("

Hello World

");
out.closr();
            }
        }
        xml的写法:(在tomcat安装目录下有很多例子)
            <web-app>
                <servlet>
                    <servlet-name>HelloWorld</servlet-name>
                    <servlet-class>first.HelloWorld</servlet-class>
                </servlet>
                <servlet-mapping>
                    <servlet-name>HelloWorld</servlet-name>
                    <url-pattern>/sayHello</url-pattern>
                </servlet-mapping>
            </web-app>

用Servlet原生写也可以啊

用原生的要怎么写,servlet太久没用了