在asp.net MVC 3中解析JSON輸出

Question

我想解析C＃asp.net mVC3中的JSON字符串，但不了解如何解析我的json字符串。 我的JSON字符串是這樣的：

{"dept":"HR","data":[{"height":5.5,"weight":55.5},{"height":5.4,"weight":59.5},{"height":5.3,"weight":67.7},{"height":5.1,"weight":45.5}]}

碼：

var allData = { dept: deptname, data: arr};
var allDataJson = JSON.stringify(allData);
$.ajax({
                url: '<%=Url.Action("funx","Controller")%>',
                data: { DJson: allDataJson },
                async: false,
                cache: false,
                type: 'POST',
                success: function (data) {
                    alert("success data: "+data);
                }

}); 

public String funx(string DJson)
{
     System.Diagnostics.Debug.WriteLine("Returned Json String:" + DJson);

     // var yourObject = new JavaScriptSerializer().Deserialize(DJson);
     return "successfull";
}

我是asp.net的新手。 我想解析該字符串並將其保存在數據庫中。

Answer 1

方法1：

使用您的JSON字符串的結構創建兩個類：

public class myobj
{
    public string dept;
    public IEnumerable<mydata>;
}

public class mydata
{
    public int weight;
    public int height;
}

然后使用它來解析它：

public static T FromJSON<T>(string str)
{
    JavaScriptSerializer serializer = new JavaScriptSerializer();
    return serializer.Deserialize<T>(str);
}

像這樣：

myobj obj = MP3_SIOP_LT.Code.Helpers.JSONHelper.FromJSON<myobj>(@"{""dept"":""HR"",""data"":[{""height"":5.5,""weight"":55.5},{""height"":5.4,""weight"":59.5},{""height"":5.3,""weight"":67.7},{""height"":5.1,""weight"":45.5}]}");

結果：

方法2：

如果您不想讓類具有JSON結構，請使用與上面相同的方法，如下所示，但是為了獲得dynamic對象：

dynamic obj = MP3_SIOP_LT.Code.Helpers.JSONHelper.FromJSON<dynamic>(@"{""dept"":""HR"",""data"":[{""height"":5.5,""weight"":55.5},{""height"":5.4,""weight"":59.5},{""height"":5.3,""weight"":67.7},{""height"":5.1,""weight"":45.5}]}");

結果：

Answer 2

首先為您的json建立模型

public class Size
{
    public double height { get; set; }
    public double weight { get; set; }
}

public class MyData
{
    public string dept { get; set; }
    public List<Size> data { get; set; }
}

現在，您可以反序列化json。 內置DataContractJsonSerializer

var serializer = new DataContractJsonSerializer(typeof(MyData));
var data = (MyData)serializer.ReadObject(new MemoryStream(Encoding.UTF8.GetBytes(json)));

或使用Json.Net

var data = JsonConvert.DeserializeObject<MyData>(json);

您甚至可以不創建任何類就采用dynamic方式

dynamic dynObj = JsonConvert.DeserializeObject(json);

foreach(var item in dynObj.data)
{
    Console.WriteLine("{0} {1}", item.height, item.weight);
}

Answer 3

您可以使用NewtonSoft Json.Net進行解析。

嘗試這個

var json = "{\"dept\":\"HR\",\"data\":[{\"height\":5.5,\"weight\":55.5},{\"height\":5.4,\"weight\":59.5},{\"height\":5.3,\"weight\":67.7},{\"height\":5.1,\"weight\":45.5}]}";
var foo = JsonConvert.DeserializeObject<RootObject>(json);

// Check Values
// var department = foo.dept;
// foreach (var item in foo.data)
// {
//      var height = item.height;
//      var weight = item.weight;
// }

public class Datum
{
    public double height { get; set; }
    public double weight { get; set; }
}

public class RootObject
{
    public string dept { get; set; }
    public List<Datum> data { get; set; }
}

Nuget： Json.Net