将 FileUpload 成功回调的对象传递给 Controller Action 方法

Question

JSON 字符串

[{"程序":"易趣 US","Date":"/Date(1384108200000)/","TimePlus":"/Date(-62135596800000)/","Campaign":"cwsi12","Clicks":0,"EPC":3.3799, “收入”：6.7599，“CampaignID”：“5337412363”，“印象”：“0”，“状态”：“重复 在数据库中"},{"程序":"eBay US","Date":"/Date(1384108200000)/","TimePlus":"/Date(-62135596800000)/","Campaign":"cwsi12","Clicks":0,"EPC":3.3799, “收入”：6.7599，“CampaignID”：“5337412363”，“印象”：“0”，“状态”：“重复 在数据库中"},{"程序":"eBay US","Date":"/Date(1384108200000)/","TimePlus":"/Date(-62135596800000)/","Campaign":"cwsi12","Clicks":0,"EPC":3.3799, “收入”：6.7599，“CampaignID”：“5337412363”，“印象”：“0”，“状态”：“重复 在数据库中"},{"程序":"eBay US","Date":"/Date(1384108200000)/","TimePlus":"/Date(-62135596800000)/","Campaign":"cwsi12","Clicks":0,"EPC":3.3799, “收入”：6.7599，“CampaignID”：“5337412363”，“印象”：“0”，“状态”：“重复 在数据库中"}]

查看模型

public class EbayEarnings_Temp
{
    public String Program { get; set; }
    public DateTime Date { get; set; }
    public DateTime TimePlus { get; set; }
    public String Campaign { get; set; }
    public int Clicks { get; set; }
    public decimal EPC { get; set; }
    public decimal Earnings { get; set; }
    public String CampaignID { get; set; }
    public String Impression { get; set; }
    public string Status { get; set; }
}

var serializer = new System.Runtime.Serialization.Json.DataContractJsonSerializer(typeof(EbayEarnings_Temp));
var c = (EbayEarnings_Temp)serializer.ReadObject(jsonString);

我收到这个错误

解析值时遇到意外字符：S. Path '', line 0, position 0.

Answer 1

消息是 json 对象而不是列表。您必须手动集成序列化程序或序列化 json 对象。 我想它一定是这样的：

public ActionResult action(Object Message)
{    
    // deserialise if Object Message is a string
    var serializer = new JavaScriptSerializer();
    var c = serializer.Deserialize<YourClass>(Message);

    // deserialise if Object Message is a JsonObject
    var serializer = new DataContractJsonSerializer(typeof(YourClass));
    var c = (YourClass)serializer.ReadObject(Message);

    return PartialView(Message);
}

或其他解决方案：

我使用库Newtonsoft.Json。如果你打算在你的情况下使用它，它看起来像：

...    
MyClass m = JsonConvert.DeserializeObject<Message>(message.ToString());
var status = m.Status; //...and so on