applicationWillBecomeActive在MonoTouch

Question

我正在嘗試使用applicationDidBecomeActive方法（和類似的方法） - 我可以在ObjC中找到大量的示例，但Monotouch中沒有。 已經嘗試了AppDelegate和UIViewController中的重寫，但編譯器找不到合適的重寫方法。那麼，我該如何使用它？ 我想使用它（與一個計時器和IdleTimerDisabled結合）來停止設備進入睡眠時間比平常更長（這是一個秒錶類型的應用程序）。也許我走錯了路。

Answer 1

在從UIApplicationDelegate繼承了您的應用程序委託，您可以重寫這些：

/// <summary> 
/// Gets called by iOS if the app is started from scratch and is not resumed from background. 
/// We have 17 seconds to leave this methos before iOS will kill the app. 
/// </summary> 
public override bool FinishedLaunching (UIApplication application, NSDictionary launchOptions) 

/// <summary> 
/// Called if the app comes back from background or gets started. Triggered after FinishedLaunching(). 
/// </summary> 
public override void OnActivated (UIApplication application) 

/// <summary> 
/// Called if the application gets pushed to the background because the user hits the home button. 
/// </summary> 
public override void DidEnterBackground (UIApplication application) 

/// <summary> 
/// Gets called if the app is resumed from background but NOT if the app starts first time. 
/// </summary> 
public override void WillEnterForeground (UIApplication application)