如何传递内容以响应 Asp.net WebAPI 中的异常过滤器？

Question

考虑以下代码：

我的问题是：

1) 我似乎无法将错误转换为 HttpContent

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

这里的示例似乎只提供 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/exception-handling

 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");

如果您想传递复杂的对象，您还可以使用CreateResponse（在RC 中添加以替换不再存在的通用HttpResponseMessage<T> 类）方法：

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