AJAX的URL怎么写能被web.xml的servlet扫描到

AJAX的URL怎么写能被web.xml的servlet扫描到，servlet怎么指向具体类具体方法，求一套完整demo

js里面 $.ajax({
url:'<%=path%>/HelloServlet',
dataType: "json",
data:{
name:"zhangsan",
password:"123123"
},
type :'GET',
success:function(data)
{
alert( $(data).get(0).name);
}
});
web.xml中

HelloServlet
com.milk.servlet.HelloServlet

HelloServlet
/HelloServlet

HelloServlet中
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException
{
System.out.println("执行了-------------");
String name = (String)request.getParameter("name");
String password = (String)request.getParameter("password");

System.out.println(name + " ," + password);

    student stu = new student();        
    stu.setAge(18);
    stu.setName("zhangsan");
    JSONArray arr=JSONArray.fromObject(stu);
    PrintWriter out = response.getWriter();
    out.println(arr.toString());    
    out.flush();
    out.close();
}

？？

向下面这样配置servlet，访问路径是localhost:8080/应用名/demo

 <servlet>  
  <servlet-name>demo</servlet-name>    
  <servlet-class>com.my.demo.system.lifeServlet</servlet-class>    
  <load-on-startup>1</load-on-startup>    
</servlet>  
<servlet-mapping>  
<servlet-name>demo</servlet-name>  
<url-pattern>/demo</url-pattern>  
</servlet-mapping>  

AJAX的URL 是由servlet的路径和Controller里的路径组合起来的，如：localhost:8080/应用名/demo/con/login对应的是

 @Controller  
 @RequestMapping("/con")
public class HelloWorldController {  

    @RequestMapping("/login")  
    public String loginFunction(){  
        return "login";  
    }  

} 

刚写的一段代码居然没了。
...


redteam

mysite.server.TeamServlet



teamColor

red





bgColor

#CC0000







redteam

/red/AccessIt



...

整套demo就没了。 就下面的将就下。 访问 http://xxxx/red/AccessIt 就能访问到 mysite.server.TeamServlet
...


redteam

mysite.server.TeamServlet



teamColor

red





bgColor

#CC0000







redteam

/red/AccessIt

...

需要配置文件设置一下，web

直接找到你需要的servlet的urlpattern 写到ajax的url上就好了