Question

所以这是我的问题，我通过ajax从控制器返回我的模型List列表。作为回应我得到List但是当我尝试用这个响应填充我的数据表时它显示未定义到处我的观点：

function GetData() {
                pid = $(this).data('id');
                var singleValues = $("#network").val();
                var id = $(this).find(":selected").val();
                var network = { "network": id };
                $.ajax({
                    type: 'POST',
                    url: '/Home/GetData',
                    contentType: 'application/json; charset=utf-8',
                    data: JSON.stringify(network),
                    success: function (response) {
                    //    console.log(response);
                        alert(response);
                            for (var i = 0; i < response.d.length; i++) {
                                $("#example").append("<tr><td>" + response.d[i].ACCOUNT_TYPE_ID + "</td><td>" + response.d[i].ACCOUNT_ISO_CODE + "</td><td>" + response.d[i].ACCOUNT_DESC + "</td></tr>");
                            debugger;
                        }
                    }

                });

MY Controller：

[HttpPost]
    public DbDataModel[] GetData(string network)
    {
        List<DbDataModel> l = new List<DbDataModel>();
        DataTable db = GetDataSource(network);
        foreach (DataRow row in db.Rows)
        {
            DbDataModel model = new DbDataModel();
            model.Account_type_id = row["ACCOUNT_TYPE_ID"].ToString();
            model.Account_iso_code = row["ACCOUNT_ISO_CODE"].ToString();
            model.Account_desc = row["ACCOUNT_DESC"].ToString();
            l.Add(model);
        }
            return l.ToArray();

    }

我的模特：

 public class DbDataModel
{
    public string Account_type_id { get; set; }
    public string Account_iso_code { get; set; }
    public string Account_desc { get; set; }
}

Answer 1

更改您的方法以返回JSON

[HttpGet]
public JsonResult GetData(string network)
{
  List<DbDataModel> l = new List<DbDataModel>();
  .....
  return Json(l, JsonRequestBehavior.AllowGet);
}

并在脚本中

...
$.ajax({
  type: 'GET',
  ...

Answer 2

这就是我做对了。

function GetData() {
                pid = $(this).data('id');
                var singleValues = $("#network").val();
                var id = $(this).find(":selected").val();
                var network = { "network": id };
                $.ajax({
                    type: 'POST',
                    url: '/Home/GetData',
                    dataType:"json",
                    traditional: true,                            
                    data: jQuery.param( network ),
                    success: function (response) {
                        debugger;
                        var oTable = $('#example').dataTable();//get the DataTable
                        oTable.fnClearTable();//clear the DataTable
                            for (var i = 0; i < response.length; i++) {
                //                $("#example").append("<tr><td>" + response[i].Account_type_id + "</td><td>" + response[i].Account_iso_code + "</td><td>" + response[i].Account_desc + "</td></tr>");
                                oTable.fnAddData([
                                    response[i].Account_type_id,
                                    response[i].Account_iso_code,
                                    response[i].Account_desc
                                ]);
                            }

                    }

                });
            };

[HttpPost]
    public ActionResult GetData(string network)
    {
        List<DbDataModel> l = new List<DbDataModel>();
        DataTable db = GetDataSource(network);
        foreach (DataRow row in db.Rows)
        {
            DbDataModel model = new DbDataModel();
            model.Account_type_id = row["ACCOUNT_TYPE_ID"].ToString();
            model.Account_iso_code = row["ACCOUNT_ISO_CODE"].ToString();
            model.Account_desc = row["ACCOUNT_DESC"].ToString();
            l.Add(model);
        }
            return Json(l.ToArray(), JsonRequestBehavior.AllowGet);            
    }

完全按照自己的意愿行事。

json在ajax中的响应未定义

2 个答案: