Asp.net MVC 4 中正确的 JSON 序列化

我建议使用ServiceStack或Json之类的东西.NET处理MVC应用程序中的Json输出.但是，您可以轻松地编写一个类，并使用基类重写Json方法.请看下面的例子.

注意:这样，您就不需要Global.ascx.cs文件中的任何内容.

Custom JsonDotNetResult class:

public class JsonDotNetResult : JsonResult
{
    private static readonly JsonSerializerSettings Settings = new JsonSerializerSettings
    {
        ContractResolver = new CamelCasePropertyNamesContractResolver(),
        Converters = new List<JsonConverter> { new StringEnumConverter() }
    };

    public override void ExecuteResult(ControllerContext context)
    {
        if (this.JsonRequestBehavior == JsonRequestBehavior.DenyGet &&
            string.Equals(context.HttpContext.Request.HttpMethod, "GET", StringComparison.OrdinalIgnoreCase))
        {
            throw new InvalidOperationException("GET request not allowed");
        }

        var response = context.HttpContext.Response;

        response.ContentType = !string.IsNullOrEmpty(this.ContentType) ? this.ContentType : "application/json";

        if (this.ContentEncoding != null)
        {
            response.ContentEncoding = this.ContentEncoding;
        }

        if (this.Data == null)
        {
            return;
        }

        response.Write(JsonConvert.SerializeObject(this.Data, Settings));
    }
}

Base Controller class:

public abstract class Controller : System.Web.Mvc.Controller
{
    protected override JsonResult Json(object data, string contentType, System.Text.Encoding contentEncoding, JsonRequestBehavior behavior)
    {
        return new JsonDotNetResult
            {
                Data = data,
                ContentType = contentType,
                ContentEncoding = contentEncoding,
                JsonRequestBehavior = behavior
            };
    }
}

现在，在你的控制器操作中，你可以简单地返回这样的内容.

return Json(myObject, JsonRequestBehavior.AllowGet);

巴姆.您现在有了随Json:)返回的CamelCase对象.

注意:对于使用Json创建的每个对象，都可以通过序列化程序设置来实现这一点.但是，谁会希望在每次返回Json时都将其输入出来呢？