AJAX异步提交form表单

比如说有如下form：

        

发送异步请求提交form：

1. function save(){

2.  

3. $.ajax({

4. url:'debt/saveNew.do'+'?t='+Math.random(),

5. data:$('#form1').serialize(), //将表单数据序列化，格式为name=value

6. type:'POST',

7. dataType:'json',

8. success:function(data){

9. //success

10. },

11. error:function(){

12. console.log("提交ajax函数异常");

13. },

14.  

15. });

16. }

 

 

 

获取（能从param中取到值意味着怎么接收都可）：

1. @RequestMapping(value = "saveNew")

2. //@Token(remove=true)

3. public void saveNew(

4. HttpServletRequest request,HttpServletResponse response){

5. response.setContentType("application/json; charset=UTF-8");

6. try {

7. String submitTime = request.getParameter("submitTime");

8. String receiverId = request.getParameter("receiverId");

9. String isRegister = request.getParameter("isRegister");