Question

我有2个HTML选择使用jQuery多选择级联绑定使用ajax调用它工作正常问题是我想在master中设置值然后绑定细节然后设置值详细但是因为它是ajax时尝试设置值详细说明尚未提供的值。

桌车 CarID int CarModel（丰田） CarSubModel（力士）

 function GetCarData()
    {
        var CarID = getQuerystringByName('C');
        if (CarID!=null)
        {
            $.ajax({
                type: "POST",
                contentType: "application/json; charset=utf-8",
                url: "Query.asmx/GetCarByID",
                data: '{CarID: ' + CarID + '}',
                dataType: "json",
                success: function (Result) {
                    Result = Result.d;
                    txtCarName.value = Result.CarName;
                    $("#ddlCarModel").val(Result.ModelID);
                      BindSubModel();// Ajax to load Sub Models
                      $("#ddlCarSubModel").val(Result.SubModelID);

我需要（某些方法）执行$（“＃ddlCarSubModel”）。val（Result.SubModelID）; 在子ajax完成后，我不想使用等待方法（定时器）。

  function BindSubModel() {
        var ModelVal = $("#ddlCarModel").val();
        if (ModelVal != "") {
                $.ajax({
            type: "POST",
            contentType: "application/json; charset=utf-8",
            url: "Query.asmx/LoadCarSubModel",
            data: '{ModelID: ' + ModelVal + '}',
            dataType: "json",
            success: function (Result) {

网络方法

     public class CarSubModelInfo
    {
        public int SubModelID { get; set; }
        public string SubModelName { get; set; }
    }

    [WebMethod]
    [ScriptMethod]
    public List<CarSubModelInfo> LoadCarSubModel(string ModelID)
    {
        string Conn = System.Configuration.ConfigurationManager.ConnectionStrings["GPS_TrackingConnectionString"].ConnectionString;

        CarSubModelInfo driver = new CarSubModelInfo();
        List<CarSubModelInfo> SubModelInformation = new List<CarSubModelInfo>();
        DataSet ds;
        using (SqlConnection con = new SqlConnection(Conn))
        {
            using (SqlCommand cmd = new SqlCommand("select * from T_SubCarModels where  ModelID=" + ModelID, con))
            {
                con.Open();
                cmd.Connection = con;
                cmd.CommandType = CommandType.Text;
                using (SqlDataAdapter da = new SqlDataAdapter(cmd))
                {

                    ds = new DataSet();
                    da.Fill(ds);
                }
            }
        }
        try
        {
            if (ds != null)
            {
                if (ds.Tables.Count > 0)
                {
                    if (ds.Tables[0].Rows.Count > 0)
                    {
                        foreach (DataRow dr in ds.Tables[0].Rows)
                        {
                            SubModelInformation.Add(new CarSubModelInfo()
                            {
                                SubModelID = Convert.ToInt32(dr["id"]),
                                SubModelName = dr["name"].ToString()
                            });
                        }
                    }
                }
            }
        }
        catch (Exception ex)
        {
            throw ex;
        }
        return SubModelInformation;
    }

Answer 1

放$("#ddlCarSubModel").val(Result.SubModelID); 在函数BindSubModel();

中取得成功

当ajax调用完成时执行脚本

1 个答案: