Question

我有telerik下拉列表

@(Html.Telerik().DropDownList().Name("InstrumentType").BindTo(Model.MITypesList).Enable(!Model.IsReadOnlyOtherProperties).HtmlAttributes(new { @class = "instrument-type" }))

左边我有按钮添加（'add-type-mi'），由ajax

保存到db

$(".add-type-mi").on("click", function (e) {
            e.preventDefault();
            var choiсeMi = $("#choiсe-type-mi");
            choiсeMi.find(".btnSubmit").one("click", function () {
                var nameId = choiсeMi.find(".className").attr('id');
                var nameType = choiсeMi.find(".className").val();

                var categoryId = choiсeMi.find(".classCategory").attr('id');
                var categoryName = choiсeMi.find(".classCategory").val();

                var strVal = [{ name: nameId, value: nameType }, { name: categoryId, value: categoryName }];

                $.ajax({
                    type: "POST",
                    url: $('#saveType').data('save-type'),
                    data: { "stringValues": JSON.stringify(strVal), "tableValues": [] },
                    cache: false,
                    dataType: 'json',
                    success: function (data) {
                    //this I need to bind data.title to dropdownlist
                        choiсeMi.data('tWindow').close();

                    }
                });
            });

            choiсeMi.find(".btnCancel").one("click", function () {
                choiсeMi.data('tWindow').close();
            });

            choiсeMi.data('tWindow').center().open();
        });

我需要将我的数据绑定到dropdownlist（'InstrumentType'）而不更新页面

Answer 1

绑定json我使用dataBind（）成功后有我的代码：function

var instrumentTypeList = $("#InstrumentType").data("tDropDownList");

        var arrayList = instrumentTypeList.data;
        if (arrayList == undefined) {
            arrayList = [];
        }

            arrayList.push({ Text: data.objFullName, Value: data.id });
            instrumentTypeList.dataBind(arrayList);

将json数据绑定到telerik下拉列表

1 个答案: