是否可以在 ASP.NET MVC 3 中使用JSON.NET作为默认 JSON 序列化程序?
根据我的研究,似乎实现这一点的唯一方法是将 ActionResult 扩展为MVC3 中的 JsonResult 不是虚拟的......
我希望通过 ASP.NET MVC 3 可以指定一个可插入的提供程序以序列化为 JSON。
想法?
是否可以在 ASP.NET MVC 3 中使用JSON.NET作为默认 JSON 序列化程序?
根据我的研究,似乎实现这一点的唯一方法是将 ActionResult 扩展为MVC3 中的 JsonResult 不是虚拟的......
我希望通过 ASP.NET MVC 3 可以指定一个可插入的提供程序以序列化为 JSON。
想法?
我相信最好的方法是 - 如您的链接中所述 - 直接扩展 ActionResult 或扩展 JsonResult 。
至于控制器上非虚的方法JsonResult 不成立,选择合适的重载即可。这很好用:
protected override JsonResult Json(object data, string contentType, Encoding contentEncoding)
编辑 1:一个 JsonResult 扩展...
public class JsonNetResult : JsonResult
{
public override void ExecuteResult(ControllerContext context)
{
if (context == null)
throw new ArgumentNullException("context");
var response = context.HttpContext.Response;
response.ContentType = !String.IsNullOrEmpty(ContentType)
? ContentType
: "application/json";
if (ContentEncoding != null)
response.ContentEncoding = ContentEncoding;
// If you need special handling, you can call another form of SerializeObject below
var serializedObject = JsonConvert.SerializeObject(Data, Formatting.Indented);
response.Write(serializedObject);
}
编辑 2:根据以下建议,我删除了对 Data 为 null 的检查。这应该使 JQuery 的较新版本感到高兴,并且似乎是明智的做法,因为响应可以被无条件地反序列化。但请注意,这不是来自 ASP.NET MVC 的 JSON 响应的默认行为,而是在没有数据时以空字符串响应。
我在不需要基本控制器或注入的情况下实现了这一点。
我使用操作过滤器将 JsonResult 替换为 JsonNetResult。
public class JsonHandlerAttribute : ActionFilterAttribute
{
public override void OnActionExecuted(ActionExecutedContext filterContext)
{
var jsonResult = filterContext.Result as JsonResult;
if (jsonResult != null)
{
filterContext.Result = new JsonNetResult
{
ContentEncoding = jsonResult.ContentEncoding,
ContentType = jsonResult.ContentType,
Data = jsonResult.Data,
JsonRequestBehavior = jsonResult.JsonRequestBehavior
};
}
base.OnActionExecuted(filterContext);
}
}
在 Global.asax.cs Application_Start() 您需要添加:
GlobalFilters.Filters.Add(new JsonHandlerAttribute());
为了完整起见,这是我从其他地方获得的 JsonNetResult 扩展类,我稍作修改以获得正确的蒸汽支持:
public class JsonNetResult : JsonResult
{
public JsonNetResult()
{
Settings = new JsonSerializerSettings
{
ReferenceLoopHandling = ReferenceLoopHandling.Error
};
}
public JsonSerializerSettings Settings { get; private set; }
public override void ExecuteResult(ControllerContext context)
{
if (context == null)
throw new ArgumentNullException("context");
if (this.JsonRequestBehavior == JsonRequestBehavior.DenyGet && string.Equals(context.HttpContext.Request.HttpMethod, "GET", StringComparison.OrdinalIgnoreCase))
throw new InvalidOperationException("JSON GET is not allowed");
HttpResponseBase response = context.HttpContext.Response;
response.ContentType = string.IsNullOrEmpty(this.ContentType) ? "application/json" : this.ContentType;
if (this.ContentEncoding != null)
response.ContentEncoding = this.ContentEncoding;
if (this.Data == null)
return;
var scriptSerializer = JsonSerializer.Create(this.Settings);
scriptSerializer.Serialize(response.Output, this.Data);
}
}
使用 Newtonsoft 的 JSON 转换器:
public ActionResult DoSomething()
{
dynamic cResponse = new ExpandoObject();
cResponse.Property1 = "value1";
cResponse.Property2 = "value2";
return Content(JsonConvert.SerializeObject(cResponse), "application/json");
}
我知道这个问题已经得到回答,但我正在使用不同的方法,因为我正在使用依赖注入来实例化我的控制器。
我已将 IActionInvoker(通过注入控制器的 ControllerActionInvoker 属性)替换为覆盖 InvokeActionMethod 方法的版本。
这意味着不会更改控制器继承,当我通过更改所有控制器的 DI 容器注册升级到 MVC4 时,可以轻松删除它
public class JsonNetActionInvoker : ControllerActionInvoker
{
protected override ActionResult InvokeActionMethod(ControllerContext controllerContext, ActionDescriptor actionDescriptor, IDictionary<string, object> parameters)
{
ActionResult invokeActionMethod = base.InvokeActionMethod(controllerContext, actionDescriptor, parameters);
if ( invokeActionMethod.GetType() == typeof(JsonResult) )
{
return new JsonNetResult(invokeActionMethod as JsonResult);
}
return invokeActionMethod;
}
private class JsonNetResult : JsonResult
{
public JsonNetResult()
{
this.ContentType = "application/json";
}
public JsonNetResult( JsonResult existing )
{
this.ContentEncoding = existing.ContentEncoding;
this.ContentType = !string.IsNullOrWhiteSpace(existing.ContentType) ? existing.ContentType : "application/json";
this.Data = existing.Data;
this.JsonRequestBehavior = existing.JsonRequestBehavior;
}
public override void ExecuteResult(ControllerContext context)
{
if (context == null)
{
throw new ArgumentNullException("context");
}
if ((this.JsonRequestBehavior == JsonRequestBehavior.DenyGet) && string.Equals(context.HttpContext.Request.HttpMethod, "GET", StringComparison.OrdinalIgnoreCase))
{
base.ExecuteResult(context); // Delegate back to allow the default exception to be thrown
}
HttpResponseBase response = context.HttpContext.Response;
response.ContentType = this.ContentType;
if (this.ContentEncoding != null)
{
response.ContentEncoding = this.ContentEncoding;
}
if (this.Data != null)
{
// Replace with your favourite serializer.
new Newtonsoft.Json.JsonSerializer().Serialize( response.Output, this.Data );
}
}
}
}
--- 编辑 - 更新以显示控制器的容器注册。我在这里使用 Unity。
private void RegisterAllControllers(List<Type> exportedTypes)
{
this.rootContainer.RegisterType<IActionInvoker, JsonNetActionInvoker>();
Func<Type, bool> isIController = typeof(IController).IsAssignableFrom;
Func<Type, bool> isIHttpController = typeof(IHttpController).IsAssignableFrom;
foreach (Type controllerType in exportedTypes.Where(isIController))
{
this.rootContainer.RegisterType(
typeof(IController),
controllerType,
controllerType.Name.Replace("Controller", string.Empty),
new InjectionProperty("ActionInvoker")
);
}
foreach (Type controllerType in exportedTypes.Where(isIHttpController))
{
this.rootContainer.RegisterType(typeof(IHttpController), controllerType, controllerType.Name);
}
}
public class UnityControllerFactory : System.Web.Mvc.IControllerFactory, System.Web.Http.Dispatcher.IHttpControllerActivator
{
readonly IUnityContainer container;
public UnityControllerFactory(IUnityContainer container)
{
this.container = container;
}
IController System.Web.Mvc.IControllerFactory.CreateController(System.Web.Routing.RequestContext requestContext, string controllerName)
{
return this.container.Resolve<IController>(controllerName);
}
SessionStateBehavior System.Web.Mvc.IControllerFactory.GetControllerSessionBehavior(RequestContext requestContext, string controllerName)
{
return SessionStateBehavior.Required;
}
void System.Web.Mvc.IControllerFactory.ReleaseController(IController controller)
{
}
IHttpController IHttpControllerActivator.Create(HttpRequestMessage request, HttpControllerDescriptor controllerDescriptor, Type controllerType)
{
return this.container.Resolve<IHttpController>(controllerType.Name);
}
}
扩展来自https://stackoverflow.com/users/183056/sami-beyoglu的答案,如果您设置 Content 类型,那么 jQuery 将能够为您将返回的数据转换为对象。
public ActionResult DoSomething()
{
dynamic cResponse = new ExpandoObject();
cResponse.Property1 = "value1";
cResponse.Property2 = "value2";
return Content(JsonConvert.SerializeObject(cResponse), "application/json");
}
我的帖子可能对某人有所帮助。
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Web;
using System.Web.Mvc;
namespace MultipleSubmit.Service
{
public abstract class BaseController : Controller
{
protected override JsonResult Json(object data, string contentType,
Encoding contentEncoding, JsonRequestBehavior behavior)
{
return new JsonNetResult
{
Data = data,
ContentType = contentType,
ContentEncoding = contentEncoding,
JsonRequestBehavior = behavior
};
}
}
}
using Newtonsoft.Json;
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Web;
using System.Web.Mvc;
namespace MultipleSubmit.Service
{
public class JsonNetResult : JsonResult
{
public JsonNetResult()
{
Settings = new JsonSerializerSettings
{
ReferenceLoopHandling = ReferenceLoopHandling.Error
};
}
public JsonSerializerSettings Settings { get; private set; }
public override void ExecuteResult(ControllerContext context)
{
if (context == null)
throw new ArgumentNullException("context");
if (this.JsonRequestBehavior == JsonRequestBehavior.DenyGet && string.Equals
(context.HttpContext.Request.HttpMethod, "GET", StringComparison.OrdinalIgnoreCase))
throw new InvalidOperationException("JSON GET is not allowed");
HttpResponseBase response = context.HttpContext.Response;
response.ContentType = string.IsNullOrEmpty(this.ContentType) ?
"application/json" : this.ContentType;
if (this.ContentEncoding != null)
response.ContentEncoding = this.ContentEncoding;
if (this.Data == null)
return;
var scriptSerializer = JsonSerializer.Create(this.Settings);
using (var sw = new StringWriter())
{
scriptSerializer.Serialize(sw, this.Data);
response.Write(sw.ToString());
}
}
}
}
public class MultipleSubmitController : BaseController
{
public JsonResult Index()
{
var data = obj1; // obj1 contains the Json data
return Json(data, JsonRequestBehavior.AllowGet);
}
}
我制作了一个版本,使 Web 服务操作类型安全且简单。你像这样使用它:
public JsonResult<MyDataContract> MyAction()
{
return new MyDataContract();
}
班上:
public class JsonResult<T> : JsonResult
{
public JsonResult(T data)
{
Data = data;
JsonRequestBehavior = JsonRequestBehavior.AllowGet;
}
public override void ExecuteResult(ControllerContext context)
{
// Use Json.Net rather than the default JavaScriptSerializer because it's faster and better
if (context == null)
throw new ArgumentNullException("context");
var response = context.HttpContext.Response;
response.ContentType = !String.IsNullOrEmpty(ContentType)
? ContentType
: "application/json";
if (ContentEncoding != null)
response.ContentEncoding = ContentEncoding;
var serializedObject = JsonConvert.SerializeObject(Data, Formatting.Indented);
response.Write(serializedObject);
}
public static implicit operator JsonResult<T>(T d)
{
return new JsonResult<T>(d);
}
}