数据表警告：无法解析来自服务器的json数据

Question

我有一个代码，代码的主要目的是实现jquery数据表并呈现json数据。我成功地从控制器以json格式获取数据。当我运行代码时，它显示一个警告json数据从服务器无法解析。请帮忙。我也附上了代码。

控制器： -

 public JsonResult This_is_final(int FirstNameCount, int LastNameCount, int TotalSalary, int GenderCount)
        {
            List<Counting> l = new List<Counting>();
            l.Add(new Counting() { FirstNamecount = FirstNameCount, LastNamecount = LastNameCount, Salary = TotalSalary, Gendercount = GenderCount });

            var todoListsResults = l.Select(
                 a => new
                 {

                     a.FirstNamecount,
                     a.LastNamecount,
                     a.Salary,
                     a.Gendercount

                 });

            var jsonData = new
            {

                aadata= todoListsResults
            };
            return Json(jsonData, JsonRequestBehavior.AllowGet);


        }

Js代码： -

function callme(fncount, lncount, i, gndr) {
    var dataset = [fncount, lncount, i, gndr];
    //var oTable = $('#example').dataTable();
    $.post('/TodoList/This_is_final', { 'FirstNameCount': fncount, 'LastNameCount': lncount, 'TotalSalary': i, 'GenderCount': gndr }, function (data) {
        alert(JSON.stringify(data));
        $('#example').dataTable({
            "aaData": data,
            "bProcessing": true,
            "bServerSide": true,
            "bDestroy": true,
               "bRetrieve": true,
            "aoColumns": [
                { "aaData": "FirstNamecount" },
                { "aaData": "LastNamecount" },
                { "aaData": "TotalSalary" },
                { "aaData": "GenderCount" }
            ]

        });

    });

}

Answer 1

使用适合您的数据格式的正确DataTables初始化代码如下所示：

$('#example').dataTable({
    "aaData": data,
    "bDestroy": true,
    "sAjaxDataProp": "aadata",
    "aoColumns": [
         { "mData": "FirstNamecount" },
         { "mData": "LastNamecount" },
         { "mData": "Salary" },
         { "mData": "Gendercount" }
    ]
});

Answer 2

将数据传递给aaData时，您需要对数据进行字符串化。

"aaData": JSON.stringify(data),