请教一个有关于json的问题

Question

/*查询基本信息不包含（联系方式的）*/
function QueryBaseInfo(btn)
{
    debugger
    $.ajax({
            url: "ResumeBaseInfo.ashx?param=QueryBaseInfo",
            data: {
                ResumeBaseInfoID:46
            },
            dataType: "json",
            type: "GET",
            success: function(data) {
                alert(data);
                $.each(data,function(i,n){
                    $("#txt_BaseInfo_Name").attr("value",n.Name),
                    $("#slt_Sex").attr("value",n.Sex),
                    $("#txt_BaseInfo_IdCard").attr("value",n.IdCard),
                    $("#txt_BaseInfo_BirthDay").attr("value",n.BirthDay),
                    $("#slt_BaseInfo_IsMarry").attr("value",n.IsMarry),
                    $("#slt_BaseInfo_WorkExpYear").attr("value",n.WorkExpYear),
                    $("#slt_BaseInfo_Salary").attr("value",n.Salary),
                    $("#slt_BaseInfo_PolitFace").attr("value",n.PolitFace),
                    $("#txt_BaseInfo_Nationality").attr("value",n.Nationality),
                    $("#txt_BaseInfo_Stature").attr("value",n.Stature),
                    $("#txt_BaseInfo_Weight").attr("value",n.Weight),
                    $("#slt_BaseInfo_LocalAreaDistrict").attr("value",n.LocalAreaDistrict),
                    $("#slt_BaseInfo_DomicileDistrict").attr("value",n.DomicileDistrict),
                    $("#slt_BaseInfo_Education").attr("value",n.Education),
                    $("#slt_BaseInfo_Degree").attr("value",n.Degree),
                    $("#slt_BaseInfo_SubMajor").attr("value",n.SubMajor),
                    $("#txt_BaseInfo_MajorName").attr("value",n.MajorName),
                    $("#slt_BaseInfo_ProfessTitle").attr("value",n.ProfessTitle),
                    $("#txt_BaseInfo_CustomizeProfessTitle").attr("value",n.CustomizeProfessTitle),
                    $("#slt_BaseInfo_ProfessGrade").attr("value",n.ProfessGrade),
                    $("#txt_BaseInfo_JobOccupName").attr("value",n.JobOccupName),
                    $("#txt_BaseInfo_OperationName").attr("value",n.OperationName),
                    $("#txt_BaseInfo_HomeTel").attr("value",n.HomeTel),
                    $("#txt_BaseInfo_CompanyTel").attr("value",n.CompanyTel),
                    $("#txt_BaseInfo_Mobile").attr("value",n.Mobile),
                    $("#txt_BaseInfo_Email").attr("value",n.Email),
                    $("#txt_BaseInfo_QQ").attr("value",n.Qq),
                    $("#txt_BaseInfo_MSN").attr("value",n.Msn),
                    $("#txt_BaseInfo_PersonBlog").attr("value",n.PersonBlog),
                    $("#txt_BaseInfo_Address").attr("value",n.Address),
                    $("#txt_BaseInfo_PostZip").attr("value",n.PostZip)  
                }); 
            }
    });
}图片内是后台返回的格式，大家帮我看看对不对？
代码是获取json数据，但是取不到，不知道是循环写得有问题，还是哪里出错了，
请大家来帮忙

Answer 1

$.each(data.Table, ... 试试看

Answer 2

$(data.Table).each(function(){
   alert(this.Name);
});

Answer 3

你定位定的不对吧  把TABLE也算进去了！ 

Answer 4

:前面键名称 把引号去掉  比如 “Table”: 换成Table:
试试

Answer 5

var dataJson = eval('('+ data + ')');
$.each(dataJson.Table...

Answer 6

以下就是后台数据转json的方法,是不是这个方法出错了？/// <summary>
    /// DataTable转Json数据
    /// </summary>
    /// <param name="dt">DataTable</param>
    /// <returns>Json数据</returns>
    private string DataTableJson(DataTable dt)
    {
        StringBuilder jsonBuilder = new StringBuilder();
        jsonBuilder.Append("{\"");
        jsonBuilder.Append(dt.TableName);
        jsonBuilder.Append("\":[");
        for (int i = 0; i < dt.Rows.Count; i++)
        {
            jsonBuilder.Append("{");
            for (int j = 0; j < dt.Columns.Count; j++)
            {
                jsonBuilder.Append("\"");
                jsonBuilder.Append(dt.Columns[j].ColumnName);
                jsonBuilder.Append("\":\"");
                jsonBuilder.Append(dt.Rows[i][j].ToString());
                jsonBuilder.Append("\",");
            }
            jsonBuilder.Remove(jsonBuilder.Length - 1, 1);
            jsonBuilder.Append("},");
        }
        jsonBuilder.Remove(jsonBuilder.Length - 1, 1);
        jsonBuilder.Append("]");
        jsonBuilder.Append("}");
        return jsonBuilder.ToString();
    }

Answer 7

问题解决了，是出在DataTable转Json函数里，谢谢大家