Question

我有一个带有动作（toggleApplication）的按钮，可以打开另一个应用程序。当我从打开的应用程序返回到我的应用程序时，我想要转到另一个视图。但是我的代码中出现以下错误：

Receiver（RootViewController：0x1f192450）没有带标识符'showReceipt'的segue

AppDelegate.m

- (BOOL)application:(UIApplication *)application openURL:(NSURL *)url sourceApplication:(NSString *)sourceApplication annotation:(id)annotation {

        RootViewController *controller = [RootViewController alloc];

        return [controller handleURL:url];

    }

RootViewController.m

- (IBAction)toggleApplication:(id)sender{

     // Open application
     [[UIApplication sharedApplication] openURL:[NSURL URLWithString:theUrlString]];

}

- (BOOL)handleURL:(NSURL *)url{

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

     return YES;

}

Answer 1

使用NSNotificationCenter计算出来。

<强> AppDelegate.m

- (BOOL)application:(UIApplication *)application openURL:(NSURL *)url sourceApplication:(NSString *)sourceApplication annotation:(id)annotation {

        [[NSNotificationCenter defaultCenter] postNotificationName:@"segueListener" object:nil];

        return YES;

    }

<强> RootViewController.m

- (void)viewDidLoad{

     [super viewDidLoad];

     [[NSNotificationCenter defaultCenter] addObserver:self selector:@selector(receiptSegue) name:@"segueListener" object:nil];

}    

- (IBAction)toggleApplication:(id)sender{

     // Open application
     [[UIApplication sharedApplication] openURL:[NSURL URLWithString:theUrlString]];

    }

   - (void)receiptSegue{
        [self performSegueWithIdentifier:@"showReceipt" sender:self];
   }

完全符合我的要求。不知道这是不是正确的做法。

在AppDelegate中的openURL之后发现

1 个答案: