如何通过Asp.net WebAPI中的异常过滤器传递内容？

Question

考虑以下代码:

我的问题是:

1)我似乎无法将错误转发给HttpContent

2)我不能使用CreateContent扩展方法,因为context.Response.Content.CreateContent上不存在

这里的例子似乎只提供了StringContent,我希望能够将内容作为JsobObject传递:http: //www.asp.net/web-api/overview/web-api-routing-and-actions/异常处理

 public class ServiceLayerExceptionFilter : ExceptionFilterAttribute
    {
        public override void OnException(HttpActionExecutedContext context)
        {
            if (context.Response == null)
            {                
                var exception = context.Exception as ModelValidationException;

                if ( exception != null )
                {
                    var modelState = new ModelStateDictionary();
                    modelState.AddModelError(exception.Key, exception.Description);

                    var errors = modelState.SelectMany(x => x.Value.Errors).Select(x => x.ErrorMessage);

                    // Cannot cast errors to HttpContent??
                    // var resp = new HttpResponseMessage(HttpStatusCode.BadRequest) {Content = errors};
                    // throw new HttpResponseException(resp);

                    // Cannot create response from extension method??
                    //context.Response.Content.CreateContent
                }
                else
                {
                    context.Response = new HttpResponseMessage(context.Exception.ConvertToHttpStatus());
                }                
            }

            base.OnException(context);
        }

    }

Answer 1

context.Response = new HttpResponseMessage(context.Exception.ConvertToHttpStatus());
context.Response.Content = new StringContent("Hello World");

如果要传递复杂对象,还可以使用CreateResponse(在RC中添加以替换HttpResponseMessage<T>不再存在的泛型类)方法:

context.Response = context.Request.CreateResponse(
    context.Exception.ConvertToHttpStatus(), 
    new MyViewModel { Foo = "bar" }
);