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);
- }
- }
解决方法
- 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中添加以替换不再存在的泛型HttpResponseMessage< T>类)方法:
- context.Response = context.Request.CreateResponse(
- context.Exception.ConvertToHttpStatus(),new Myviewmodel { Foo = "bar" }
- );