在asp.net MVC 3中解析JSON输出
在asp.net MVC 3中解析JSON输出
提问于 2015-08-20 06:30:30
我想在C# asp.net mVC3中解析JSON字符串，但不知道如何解析json字符串。我的JSON字符串是这样的：
{"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}]}代码：
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。我想解析该字符串并将其保存在数据库中。
回答 3
Stack Overflow用户
回答已采纳
发布于 2015-08-20 06:42:55
首先，为json创建一个模型。
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; }
}现在您可以反序列化您的json。内置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);您甚至可以使用dynamic方式，而无需创建任何类。
dynamic dynObj = JsonConvert.DeserializeObject(json);
foreach(var item in dynObj.data)
{
Console.WriteLine("{0} {1}", item.height, item.weight);
}Stack Overflow用户
发布于 2015-08-20 06:42:13
方法1:
使用JSON字符串的结构创建两个类：
public class myobj
{
public string dept;
public IEnumerable<mydata>;
}
public class mydata
{
public int weight;
public int height;
}在此之后，使用这个来解析它：
public static T FromJSON<T>(string str)
{
JavaScriptSerializer serializer = new JavaScriptSerializer();
return serializer.Deserialize<T>(str);
}如下所示：
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}]}");结果：
方法2:
如果您不想拥有带有JSON结构的类，请使用与上面类似的方法，但是为了获得一个dynamic对象：
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}]}");结果：
Stack Overflow用户
发布于 2015-08-20 06:43:39
您可以使用NewtonSoft Json.Net进行解析。
尝尝这个
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
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/32111018
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