AJAX异步提交form表单

记录：

网上有说怎么做，没说怎么接收，打印了一下数据，记录一下取值:

比如说有如下form：

<form id="form1" name="form1" action="" method="post">
            <input type="hidden" name="submitTime" value="${now }">
            <input type="hidden" name="receiverId" value="${receiverID }">
            <input type="hidden" name="isRegister" value="${isRegister }">
</form>  

       

发送异步请求提交form：

function save(){

    $.ajax({
        url:'debt/saveNew.do'+'?t='+Math.random(),
        data:$('#form1').serialize(),                 //将表单数据序列化，格式为name=value
        type:'POST',
        dataType:'json',
        success:function(data){
        //success
        },
        error:function(){
            console.log("提交ajax函数异常");
        },
        
    });
}

获取（能从param中取到值意味着怎么接收都可）：

@RequestMapping(value = "saveNew")
    //@Token(remove=true)
    public void saveNew(
            HttpServletRequest request,HttpServletResponse response){
        response.setContentType("application/json; charset=UTF-8");
        try {
            String submitTime = request.getParameter("submitTime");
            String receiverId = request.getParameter("receiverId");
            String isRegister = request.getParameter("isRegister");