asp.net webform javascript postback JSON

在弹出界面点击按钮触发后台的postback而不是刷新界面（保留已存在的搜索条件）

function filterData() {
    var data = { col: $('#filterPopup').attr('name'), val: $('#filterPopupInput').val() };
    __doPostBack('filterColumn', JSON.stringify(data));
}

if (IsPostBack) {
    if (Request.Params["__EVENTTARGET"] != null && Request.Params["__EVENTTARGET"] == "filterColumn")
    {
        string data = Request["__EVENTARGUMENT"];
        JavaScriptSerializer jss = new JavaScriptSerializer();
        var d = jss.Deserialize<dynamic>(data) as Dictionary<string, object>;
        string filterColumn = d["col"].ToString();
        string filgterValue = d["val"].ToString();
    }
}