8

我正在尝试创建一个触发发生的系统,因此门打开 5 秒钟,然后再次关闭。我为此使用 Threading.Timer,使用:

OpenDoor();
System.Threading.TimerCallback cb = new System.Threading.TimerCallback(OnTimedEvent);
_timer = new System.Threading.Timer(cb, null, 5000, 5000);
...
void OnTimedEvent(object obj)
{
    _timer.Dispose();
    log.DebugFormat("All doors are closed because of timer");
    CloseDoors();
}

当我打开某扇门时,计时器启动。5 秒后,一切再次关闭。

但是当我打开某扇门时,等待 2 秒,然后打开另一扇门,3 秒后一切都关闭。如何“重置”计时器?

4

2 回答 2

14

不要丢弃计时器,每次打开门时都要更换它,例如

// Trigger again in 5 seconds. Pass -1 as second param to prevent periodic triggering.
_timer.Change(5000, -1); 
于 2013-06-14T13:19:44.403 回答
7

你可以这样做:

// First off, initialize the timer
_timer = new System.Threading.Timer(OnTimedEvent, null,
    Timeout.Infinite, Timeout.Infinite);

// Then, each time when door opens, start/reset it by changing its dueTime
_timer.Change(5000, Timeout.Infinite);

// And finally stop it in the event handler
void OnTimedEvent(object obj)
{
    _timer.Change(Timeout.Infinite, Timeout.Infinite);
    Console.WriteLine("All doors are closed because of timer");
}
于 2013-06-14T13:21:22.047 回答