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");

巴特西

AJAX异步提交form表单

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