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.
