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我正在为 jqGrid 使用强类型帮助器 - Lib.Web.Mvc

我不能强制 jqGrid 对我的数据进行排序。当我单击标题中的箭头时，我在 Firebug 中收到以下错误：

未捕获的异常：jqGrid - 没有这样的方法：resetSelection

当我点击行中的单元格时，我得到：

未捕获的异常：jqGrid - 没有这样的方法：setSelection

这是我的 js 代码：

@{
      var grid = new Lib.Web.Mvc.JQuery.JqGrid.JqGridHelper<ViolationViewModel>("products",
      dataType: Lib.Web.Mvc.JQuery.JqGrid.JqGridDataTypes.Json,
      methodType: Lib.Web.Mvc.JQuery.JqGrid.JqGridMethodTypes.Post,
      pager: true,
      rowsNumber: 5,
      sortingName: "Id",
      sortingOrder: Lib.Web.Mvc.JQuery.JqGrid.JqGridSortingOrders.Asc,
      url: Url.Action("Violation", "Cabinet"),
      viewRecords: true,
      rowsList: new List<int>() { 5, 10, 20, 30, 50, 100 },
      autoWidth: true,
      loadOnce: true,
      rowsNumbers: true
    );
}

我正在使用loadOnce选项。

这段代码在行动：

        [AcceptVerbs(HttpVerbs.Post)]
        public ActionResult Violation(JqGridRequest request)
        {
            JqGridResponse response = new JqGridResponse()
                {
                    TotalRecordsCount = 7
                };
            response.Records.Add(new JqGridRecord<ViolationViewModel>("1", new ViolationViewModel
                {
                    Id = 1,
                    Name = "Test1"
                }));

            response.Records.Add(new JqGridRecord<ViolationViewModel>("2", new ViolationViewModel
            {
                Id = 2,
                Name = "Test2"
            }));

            response.Records.Add(new JqGridRecord<ViolationViewModel>("3", new ViolationViewModel
            {
                Id = 3,
                Name = "Test3"
            }));

            response.Records.Add(new JqGridRecord<ViolationViewModel>("4", new ViolationViewModel
            {
                Id = 4,
                Name = "Test4"
            }));

            response.Records.Add(new JqGridRecord<ViolationViewModel>("5", new ViolationViewModel
            {
                Id = 5,
                Name = "Test5"
            }));

            response.Records.Add(new JqGridRecord<ViolationViewModel>("6", new ViolationViewModel
            {
                Id = 6,
                Name = "Test6"
            }));

            response.Records.Add(new JqGridRecord<ViolationViewModel>("7", new ViolationViewModel
            {
                Id = 7,
                Name = "Test7"
            }));

            return new JqGridJsonResult() { Data = response };
        }
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