Question

我有一个使用推送通知的代码，但我想开始调试我的代码的一些功能，并希望在模拟器中运行，我的问题是

如何告诉xcode忽略didregisterForRemote ....

我可以在此基础上评论和发挥什么？

任何帮助将不胜感激

谢谢

- (void)application:(UIApplication*)application didRegisterForRemoteNotificationsWithDeviceToken:(NSData*)deviceToken {
    NSString *device_token =[[[[deviceToken description] stringByReplacingOccurrencesOfString:@"<"withString:@""]
                                                          stringByReplacingOccurrencesOfString:@">" withString:@""]
                                                          stringByReplacingOccurrencesOfString: @" " withString: @""];
    [self checkUUID];
    Authentication *auth = [[Authentication alloc] initWithObject];
    int erro = [auth insertDeviceInfo:device_token];

    UINavigationController *navigationController = [self customizedNavigationController];
    self.viewController = [[ViewController alloc] initWithNibName:@"ViewController" bundle:nil];
    self.viewController.erro = erro;
    [navigationController setViewControllers: [NSArray arrayWithObject: self.viewController]];
    [self setNavigationController:navigationController];

    [self.window setRootViewController: navigationController];
    [self.window makeKeyAndVisible];
}

我只是希望能够使用模拟器来移动我的调试

Answer 1

最好的方法是使用宏TARGET_IPHONE_SIMULATOR

#if !(TARGET_IPHONE_SIMULATOR)
 NSLog(@"this is only real device code");
#endif

Answer 2

如果您想使用编码实现该目标，请在

中使用以下条件

didRegisterForRemoteNotificationsWithDeviceToken：

在模拟器上忽略此类功能的方法：

- （void）application：（UIApplication *）application didRegisterForRemoteNotificationsWithDeviceToken：（NSData *）deviceToken {
 if ([model isEqualToString:@"iPhone Simulator"]) 
 {
    //device is simulator
 }
 else
 {
   //Actual Device
 }
}

Xcode - 模拟器 - 推送通知

2 个答案: