Question

我制作了一个jquery过滤器工具，它以json的形式返回我的服务器端过滤数据。我想将它转换为C＃类，我也想将任何C＃类转换为我的json。

我的json和它的镜像C＃类如下：

[{"field":{"label":"Category","value":"category"},"operator":{"label":"any of","value":"in"},"value":{"label":"(Family, Friends)","value":"1,2"}},{"field":{"label":"Age","value":"age"},"operator":{"label":">","value":"gt"},"value":{"label":"18","value":"18"}},{"field":{"label":"Firstname","value":"firstname"},"operator":{"label":"equals","value":"eq"},"value":{"label":"\"test\"","value":"test"}},{"field":{"label":"Lastname","value":"lastname"},"operator":{"label":"equals","value":"eq"},"value":{"label":"\"test2\"","value":"test2"}}]

C＃镜像：



public class Field
{
    public string label { get; set; }
    public string value { get; set; }
}

public class Operator
{
    public string label { get; set; }
    public string value { get; set; }
}

public class Value
{
    public string label { get; set; }
    public string value { get; set; }
}

public class RootObject
{
    public Field field { get; set; }
    public Operator @operator { get; set; }
    public Value value { get; set; }
}

我试过这种方式：

   public class ViewFilter
{
    public List<Field> Fields { get; set; }
    public List<Operator> Operators { get; set; }
    public List<Value> Values { get; set; }
    public List<RootObject> RootObjects { get; set; }
}
    public class Field
    {
        public string label { get; set; }
        public string value { get; set; }
    }

    public class Operator
    {
        public string label { get; set; }
        public string value { get; set; }
    }

    public class Value
    {
        public string label { get; set; }
        public string value { get; set; }
    }

    public class RootObject
    {
        public Field field { get; set; }
        public Operator @operator { get; set; }
        public Value value { get; set; }
    }

我试过了：

 var result =  JsonConvert.DeserializeObject<List<ViewModel.ViewFilter>>(filter).ToList();


       foreach (ViewModel.ViewFilter item in result)
       {


       }

Answer 1

接收数据时，请使用NewtonSoft JsonConvert类：

var serializerSettings = new JsonSerializerSettings
    { ContractResolver = new CamelCasePropertyNamesContractResolver() };
var fields = JsonConvert.DeserializeObject<List<RootObject>>(yourString, serializerSettings);

yourString 值是您获取的包含json数据的字符串。

CamelCasePropertyNamesContractResolver负责序列化驼峰大小写中的对象名称，如其名称所示。

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1 个答案: