Ajax提交信息，參數過多怎么辦？

1、以Json格式傳給后台的Controller

var data = {
        "UserId": Uid,
        "QID": Qid,
        "Age": age,
        "Sex": sex,
    };
var json = "[" + JSON.stringify(data) + "]";
 $.ajax({
        url: "/Home/User",
        type: 'post',
        data: { data: json },
        cache: false,
        success: function (text) {
           Console.log(text);
        },
        error: function (jqXHR, textStatus, errorThrown) {
            alert(jqXHR.responseText);

        }
    });

 

后台Controller接收

    public ActionResult UpdateQuestion(string data)
    {
       var question= = (JArray)Newtonsoft.Json.JsonConvert.DeserializeObject<user>(data);
    }

2、使用new FormData

2.1 HTML頁面代碼

 <form action="/Home/User" id="SubmitForm" method="post" class="form-horizontal" enctype="multipart/form-data">
            
       <div class="submit_bar">
           <input type="button" class="btn" name="" value="提交">
        </div>
</form>

2.2 JS代碼,當點擊“提交”按鈕時觸發

            var from = new FormData($("#SubmitForm")[0]);
            $.ajax({
                type: "POST",
                url: "/Home/User",
                data: from,
                processData: false,//必須為 false
                contentType: false,//必須為 false
                success: function (data) {
                    Console.log(data);
                },
                Error: function (e) {
                    console.log(e);
                }
            });

2.3 后台代碼

[HttpPost]
public async Task<IActionResult> User(UserDTO Dto)
{

}