ajax如何实现前后台简单的数据交互

<%@ page language="java" contentType="text/html; charset=UTF-8"
    pageEncoding="UTF-8"%>











	

		

			

				
用 户 名：
				

					
				
			
			

				
密　　码：
				

					
				
			
				
			

				
确认密码：
				

					
				
			
				
			

				

					

				
			
	
		
	

	

                @ResponseBody
		@RequestMapping(value = "ajaxdemo" , method = RequestMethod.POST)
		public  JSONObject  ajaxdemo(HttpServletRequest req){
		//	System.out.println(name);
		JSONObject jsonObject = new JSONObject();
		jsonObject.put("demo", "demo");
		System.out.println("-----");
		
		String parameter = req.getParameter("name");
		System.out.println(parameter);
		String parameter2 = req.getParameter("password");
		System.out.println(parameter2);

		
	
			
			
			return jsonObject;
			
		}

@RequestBody

	@ResponseBody
		@RequestMapping(value = "ajaxdemo" , method = RequestMethod.POST)
		public  JSONObject  ajaxdemo(@RequestBody String req){
			System.out.println(req);
		JSONObject jsonObject = new JSONObject();
		jsonObject.put("demo", "demo");
		System.out.println("-----");
		
//		String parameter = req.getParameter("name");
//		System.out.println(parameter);
//		String parameter2 = req.getParameter("password");
//		System.out.println(parameter2);

		
	
		
			
			return jsonObject;
			
		}