Question

任何人都可以提供帮助，下面的代码在应用程序启动时从纵向视图开始工作，如果我旋转到横向如果效果很好并显示新视图，但是当旋转回到肖像时，而不是获取原始视图，我只是得到一次崩溃，非常感谢

`

- (void)viewDidAppear:(BOOL)animated
     {
     [super viewDidAppear:animated];

     [self enableRotation];

     }

     - (void)didRotate
    {
    [self setNeedsStatusBarAppearanceUpdate];
    if ([UIDevice currentDevice].orientation == UIDeviceOrientationLandscapeLeft ||
    [UIDevice currentDevice].orientation == UIDeviceOrientationLandscapeRight) {

    [self performSegueWithIdentifier: @"segueToNumbers"sender: self];

    }

     else if ([UIDevice currentDevice].orientation == UIDeviceOrientationPortrait ||
         [UIDevice currentDevice].orientation == UIDeviceOrientationPortraitUpsideDown){

    [self performSegueWithIdentifier: @"segueToChoose"sender: self];

    }

    }


     - (void)enableRotation
    {
   [[UIDevice currentDevice] beginGeneratingDeviceOrientationNotifications];
   [[NSNotificationCenter defaultCenter] addObserver:self selector:@selector(didRotate)
                                             name:UIDeviceOrientationDidChangeNotification
                                           object:nil];
   }

`

Answer 1

不要回应旋转事件。您需要使用AutoLayout设计UI以适应不同的屏幕大小。在iOS 8中，方向是无关紧要的，您的代码应该考虑大小类

segue启用旋转

1 个答案: