javascript - 使用 ajax 将具有值的对象从 View 传递到 ActionResult

Question

我想将对象从 View 发送到 Controller ，ajax 命中操作结果但数据未传递到操作结果

查看:

 function submit()
{
    var obj = {};
    obj.PD = getResult('pd');
    obj.MD = getResult('md');
    obj.C = getResult('c');
    obj.ME = getResult('me');
    obj.EE = getResult('ee');
    obj.SED = getResult('sed');
    obj.RT = getResult('rt');
    obj.SEA = getResult('sea');
    $.ajax({
        url: '/Assessment/AssessNow',
        type: 'POST',
        async: false,
        data: '{obj' + JSON.stringify(obj) + '}',
        dataType: 'json',
        success: function (res) {

        },
        error: function (msg) {
        }
    });
    //alert(getResult('pd'));
}

型号:

 public class QAViewModel
    {
        public string C { get; set; }
        public string EE { get; set; }
        public string MD { get; set; }
        public string ME { get; set; }
        public string PD { get; set; }
        public string RT { get; set; }
        public string SEA { get; set; }
        public string SED { get; set; }
    }

Controller :

Answer 1

编辑作为一个好的观点被提出:

在帖子中你可以像这样传递完整的对象:

 function submit()
{
    var obj = {};
    obj.PD = getResult('pd');
    obj.MD = getResult('md');
    obj.C = getResult('c');
    obj.ME = getResult('me');
    obj.EE = getResult('ee');
    obj.SED = getResult('sed');
    obj.RT = getResult('rt');
    obj.SEA = getResult('sea');
    $.ajax({
        url: '/Assessment/AssessNow',
        type: 'POST',
        async: false,
        data: obj,
        success: function (res) {

        },
        error: function (msg) {
        }
    });
    //alert(getResult('pd'));
}

如果你想坚持使用 json，那么相应地修改你的 ajax 调用(你的错误在于你构建数据属性的方式:

                $.ajax({
                    url: '/Assessment/AssessNow',
                    type: 'POST',
                    async: false,
                    data: JSON.stringify({obj: obj}),
                    contentType: 'application/json; charset=utf-8',
                    dataType: 'json',
                    success: function (res) {

                    },
                    error: function (msg) {
                    }
                });

此外，根据您对结果所做的操作，您可能需要翻转 Controller 操作以返回 JsonResult(如果您正在执行某些操作，例如返回部分 View 以进行加载，请忽略):

    [HttpPost]
    public JsonResult Whatever(QAViewModel obj)
    {
        return Json(whatever, JsonRequestBehavior.AllowGet);
    }