请考虑以下代码:
我的问题是:
1)我似乎无法将错误转换为HttpContent
2)我无法使用CreateContent扩展方法,因为context.Response.Content.CreateContent
上不存在此处的示例似乎只提供了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/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);
}
}
答案 0 :(得分:13)
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通用HttpResponseMessage<T>
类)方法:
context.Response = context.Request.CreateResponse(
context.Exception.ConvertToHttpStatus(),
new MyViewModel { Foo = "bar" }
);