2013-09-24 110 views
-3

下面一个是我的JSON对象如何为json对象创建c#类?

{ 
    "LNK" : "862", 
    "MUK" : "861", 
    "query" : "java" 
} 

我想捕捉LNK和相应的处理。我已经试过:

var result = Deserializejson<Dictionary<string,Search>>(jsonresponse); 
var searchhandles = new List<Search>(result.Count); 
+1

你试过了什么代码? – Chris

+0

var result = Deserializejson >(jsonresponse); var searchhandles = new List (result.Count); –

+0

{“be”:“比利时”,“fr”:“法国”,“英国”:“大不列颠”,“nl”:“荷兰”}。我想在C#List中捕获上面的键值对 –

回答

1

如果使用JSON.Net和Deserilize的JSON为目标

dynamic obj= JsonConvert.DeserializeObject<dynamic>(content); 
1

创建一个模型类映射到您的JSON Sructure或使用dynmamics

:编辑:对不起语法是错误的,这里我们有一个完整的例子

 var seri = JsonSerializer.CreateDefault(); 

     dynamic dynamicObject = new 
     { 
      LNK = "862", 
      MUK = "861", 
      query = "java" 
     }; 

     using (var textWriter = new StringWriter()) 
     { 
      var writer = new JsonTextWriter(textWriter); 
      seri.Serialize(writer, dynamicObject); 

      var result = textWriter.ToString(); 
     } 

也工作:

 dynamic dynamicObject = new 
     { 
      LNK = "862", 
      MUK = "861", 
      query = "java" 
     }; 

     var result = JsonConvert.SerializeObject(dynamicObject); 

对于序列化/反序列化,有不同的库。除了.Net标准外,我还强烈建议通过Nuget提供Json.Net库。