相关疑难解决方法(0)

这个问题已被多次询问,但我找到的解决方案似乎都没有起作用.让我觉得这可能是一个新问题,也许是ASP.NET MVC 3的特定内容.

我正在使用JQuery Ajax对ASP.NET MVC 3控制器进行简单调用.像这样

var addressInfo = { 
    Address1: "423 Judy Road",
    Address2: "1001",
    City: "New York",
    State: "NY",
    ZipCode: "10301",
    Country: "USA" 
};

$.ajax({
    url: '/home/check',
    type: 'POST',
    data: JSON.stringify(addressInfo),
    dataType: 'json',
    contentType: 'application/json; charset=utf-8',
    success: function () {
        alert("success");
    },
    error: function () {
        alert("error");
    }
});

控制器看起来像这样

[HttpPost]
public ActionResult Check(AddressInfo addressInfo)
{
    // Do something and return Json(something)
}

这不起作用,虽然它应该基于Scottgu的帖子的"JavaScript和AJAX改进"部分

我尝试了很多不同的变体,例如:

var args = new Object();
args.addressInfo = { 
    Address1: "423 …

我正在研究以CSV格式导出数据的机制.我JSON使用jQuery以格式发送数据:

var data = JSON.stringify(dataToSend);
 $.post('DumpToCSV', { 'data': data });

然后在控制器中我生成一个CSV文件:

 public ActionResult  DumpToCSV(string data)
    {
        Response.Clear();

        XmlNode xml = JsonConvert.DeserializeXmlNode("{records:{record:" + data + "}}");

        XmlDocument xmldoc = new XmlDocument();
        //Create XmlDoc Object
        xmldoc.LoadXml(xml.InnerXml);
        //Create XML Steam 
        var xmlReader = new XmlNodeReader(xmldoc);
        DataSet dataSet = new DataSet();
        //Load Dataset with Xml
        dataSet.ReadXml(xmlReader);
        //return single table inside of dataset
        var csv = CustomReportBusinessModel.ToCSV(dataSet.Tables[0], ",");

        HttpContext context = System.Web.HttpContext.Current;
        context.Response.Write(csv);
        context.Response.ContentType = "text/csv";
        context.Response.AddHeader("Content-Disposition", …

Jquery ajax post请求将null json对象发布到mvc控制器.知道为什么会这样吗？

干杯

这是我的模特

public class CommentModel
    {
       public string EmailAddress { get; set; }
       public string Name { get; set; }
       public int ActivityId { get; set; }
       public string CommentText { get; set; }

    }

调节器

 [HttpPost]
        public ActionResult Index(CommentModel commentModel)
        {

            int i = commentModel.ActivityId;
            string k = commentModel.CommentText;

          return View();
        }

JQuery的

$("#CommentForm").submit(function () {

        var formDataAsJson = GetFormDataAsJson();

        $.ajax({
            url: $(this).attr("action"),
            dataType: 'json',
            type: "POST",
            data: JSON.stringify({ commentModel: formDataAsJson }),
            contentType: 'application/json; charset=utf-8', …