我有这个……
- public void FooAsync()
- {
- AsyncManager.OutstandingOperations.Increment();
- Task.Factory.StartNew(() =>
- {
- try
- {
- doSomething.Start();
- }
- catch (Exception e)
- {
- AsyncManager.Parameters["exc"] = e;
- }
- finally
- {
- AsyncManager.OutstandingOperations.Decrement();
- }
- });
- }
- public ActionResult FooCompleted(Exception exc)
- {
- if (exc != null)
- {
- throw exc;
- }
- return View();
- }
有没有更好的方法将异常传递回ASP.net？
干杯,伊恩.
解决方法
任务将为您捕获例外情况.如果调用task.Wait(),它将在AggregateException中包装任何捕获的异常并抛出它.
- [HandleError]
- public void FooAsync()
- {
- AsyncManager.OutstandingOperations.Increment();
- AsyncManager.Parameters["task"] = Task.Factory.StartNew(() =>
- {
- try
- {
- DoSomething();
- }
- // no "catch" block. "Task" takes care of this for us.
- finally
- {
- AsyncManager.OutstandingOperations.Decrement();
- }
- });
- }
- public ActionResult FooCompleted(Task task)
- {
- // Exception will be re-thrown here...
- task.Wait();
- return View();
- }
简单地添加[HandleError]属性是不够的.由于异常发生在不同的线程中,我们必须将异常返回到ASP.NET线程,以便对它执行任何操作.只有在我们从正确的位置抛出异常之后,[HandleError]属性才能完成它的工作.
