Question

我正在尝试创建以下JSON数据：

{
'chart.labels': ['Bob','Lucy','Gary','Hoolio'],
'chart.tooltips': ['Bob did well',
               'Lucy had her best result',
               'Gary - not so good',
               'Hoolio had a good start'
              ]
}

我正在使用C＃并尝试创建一个对象以便执行此操作......类似于：

public class chart{
 public string[] chart.labels {get;set;}
 public string[] chart.tooltips {get;set;}
}

但显然我不能拥有包含空格的属性。

我将如何做到这一点？

更新

使用JamieC的答案，以下作品完美

public virtual ActionResult CompanyStatus()
    {
        var labelList = new List<string>() { "Bob", "Lucy", "Gary", "Hoolio" };
        var tooltipsList = new List<string>() { "Bob did well", "Lucy had her best result", "Gary - not so good", "Hoolio had a good start" };

        var cData = new chartData()
        {
            Labels = labelList.ToArray(),
            Tooltips = tooltipsList.ToArray()
        };

        var serializer = new DataContractJsonSerializer(cData.GetType());
        String output;
        using (var ms = new MemoryStream())
        {
            serializer.WriteObject(ms, cData);
            output = Encoding.Default.GetString(ms.ToArray());
        } 

        return this.Content(output);
    }


    [DataContract]
    public class chartData
    {
        [DataMember(Name = "chart.labels")]
        public string[] Labels { get; set; }

        [DataMember(Name = "chart.tooltips")]
        public string[] Tooltips { get; set; }

    }
}

产生：

{"chart.labels":["Bob","Lucy","Gary","Hoolio"],"chart.tooltips":["Bob did well","Lucy had her best result","Gary - not so good","Hoolio had a good start"]}

Answer 1

通常的方法是使用DataContractJsonSerializer将对象转换为Json，并使用DataMember属性来注释要用于属性的名称：

[DataContract]
public class ChartModel{
 [DataMember(Name = "chart.labels")]
 public string[] Labels {get;set;}
 [DataMember(Name = "chart.tooltips")]
 public string[] Tooltips {get;set;}
}

我个人使用自己的ActionResult来完成MVC中的序列化：

public class JsonDataContractResult : ActionResult
{
    public JsonDataContractResult(Object data)
    {
        this.Data = data;
    }

    protected JsonDataContractResult()
    {

    }

    public Object Data { get; private set; }

    public override void ExecuteResult(ControllerContext context)
    {
        Guard.ArgumentNotNull(context, "context");

        var serializer = new DataContractJsonSerializer(this.Data.GetType());
        String output; 
        using (var ms = new MemoryStream())
        {
            serializer.WriteObject(ms, this.Data);
            output = Encoding.Default.GetString(ms.ToArray());
        } 
        context.HttpContext.Response.ContentType = "application/json";
        context.HttpContext.Response.Write(output);
    }        
}

从基本控制器中的辅助方法返回：

public abstract class MyBaseController: Controller
{

    protected JsonDataContractResult JsonContract(Object data)
    {
        return new JsonDataContractResult(data);
    }

}

然后我的控制器变得非常简单：

public class SomeController: MyBaseController
{
    public ActionResult SomeAction()
    { 
        var model = new ChartModel()
                   { 
                       Labels = ..., 
                       Tooltips = ... 
                   };
        return JsonContract(model);
    }
}

Answer 2

对于MVC项目Newtonsoft.Json库是可用的。（您必须手动将其包含在其他项目中）

所以在模型中给出JsonProperty，就像这样......

public class ChartModel{
    [JsonProperty("chart.labels")]
    public string[] Labels {get;set;}
    [JsonProperty("chart.tooltips")]
    public string[] Tooltips {get;set;}
}

并使用Newtonsoft.Json.JsonConvert.SerializeObject(object);或Json.Encode(object)转换为JSON。

使用C＃Object创建JSON

2 个答案: