尝试呈现其视图不在窗口层次结构中的UIViewController

1 回答

• 2

  您需要获取根视图控制器并从该视图控制器执行segue . 这对于调试来说非常令人沮丧，但是这里有关于这个主题的一些答案 .

  以下是我在应用程序收到推送通知时用于从根视图控制器执行segue到屏幕的一些代码 .

  UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"YourStoryboard" bundle:nil];
  
  YourViewController *viewController = [storyboard instantiateViewControllerWithIdentifier:@"viewcontrolleridentifier"];
  
  UIViewController *top = [UIApplication sharedApplication].keyWindow.rootViewController;
  
  [top presentViewController:viewController animated:YES completion: nil];
  

  在swift中使用相同的代码：

  let storyboard = UIStoryboard.init(name: "YourStoryboard", bundle: nil)
  
  let viewController = storyboard.instantiateViewController(withIdentifier: "viewcontrolleridentifier")
  
  let top = UIApplication.shared.keyWindow?.rootViewController
  
  top?.present(viewController, animated: true, completion: nil)
  

  确保在故事板中设置视图控制器标识符 .

  编辑*如果您正在访问的视图控制器嵌入在导航控制器中，您将需要修改上述代码，

  目标C：

  UIViewController *top = [self topMostController];
  
  UINavigationController *navigationController = [[UINavigationController alloc] initWithRootViewController:viewController];
  
  [top presentViewController:navigationController animated:YES completion: nil];
  

  迅速：

  let top = UIApplication.shared.keyWindow?.rootViewController
  
  let navigationController = UINavigationController.init(rootViewController: viewController)
  
  top?.present(navigationController, animated: true, completion: nil)
  

  回复于 2024-04-20T22:16:34+08:00