按下菜单按钮后如何将 Apple TV 应用程序置于后台

Question

我尝试在按下菜单按钮后使用私有方法将应用程序置于后台；并且以下代码可以正常工作：

 @implementation ViewController {
     UITapGestureRecognizer *tapRecognizer;
  }

 -(void)viewDidLoad {
     [super viewDidLoad];

     tapRecognizer = [[UITapGestureRecognizer alloc]initWithTarget:self action:@selector(handleTap:)];
     tapRecognizer.allowedPressTypes = @[[NSNumber numberWithInteger:UIPressTypeMenu]];
     [self.view addGestureRecognizer:tapRecognizer];
 }

 -(void)handleTap:(UITapGestureRecognizer *)sender {
     if (sender.state == UIGestureRecognizerStateEnded) {
         NSLog(@"Menu button released");
         UIApplication *app = [UIApplication sharedApplication];
         [app performSelector:@selector(suspend)];
     }
 }

但我想在不使用私有方法“suspend”的情况下获得相同的结果，因为 Apple 可能会因为使用私有方法而拒绝该应用程序。

 [app performSelector:@selector(suspend)];

任何建议将不胜感激

Answer 1

Swift 3 解决方案：

override func viewDidLoad() {
    super.viewDidLoad()

    let tapRecognizer = UITapGestureRecognizer(target: self, action: #selector(handleTap(gesture:)))
    tapRecognizer.allowedPressTypes = [NSNumber(value: UIPressType.menu.rawValue)]
    self.view.addGestureRecognizer(tapRecognizer)
}

func handleTap(gesture: UITapGestureRecognizer){
    if gesture.state == UIGestureRecognizerState.ended {
        let app = UIApplication.shared
        app.perform(#selector(URLSessionTask.suspend))
    }
}