Question

我们正在开发iPhone应用程序，并且我们已经实现了推送通知功能。当应用程序处于非活动状态时，收到的任何通知都会显示警报，并且徽章图标计数器会递增。当用户点击应用程序的图标或警报时，应用程序将变为活动状态，一切正常。

我们想要的是：在新通知中，每当用户点击应用图标时，它都应该打开特定页面（例如通知页面）。目前，当应用程序变为活动状态时，它会显示上一个先前打开的页面。

如何将应用程序（在前台）重定向到特定页面？

Answer 1

假设这是您的PayLoad

{"aps":{"alert":"Hai User","type":"center","badge":1,"sound":"default"}}

didReceiveRemoteNotification 方法处理有效负载

-(void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo
{
    // this is used for clear the count when user press the notification.

    [[UIApplication sharedApplication] setApplicationIconBadgeNumber:0];
    [[UIApplication sharedApplication] cancelAllLocalNotifications];


  //When notification is pressed on background it comes here

  //Get strings based on information on your json payload for example

    NSMutableString *notificationType =[[userInfo objectForKey:@"aps"]objectForKey:@"type"];

  if([notificationType isEqualToString:@"center"]){
       //redirect/push a viewcontroler here , it automatically navigate to your VC.

       profileViewController *profile = [[profileViewController alloc] initWithNibName:@"profile ViewController" bundle:nil];

    [self.window.rootViewController presentViewController: profile animated:YES completion:NULL];
  }
}

来自Apple的

通知类型为here

Answer 2

在你的

里面

- (void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo { [self openNewViewController]; // read the userInfo from the payload and based on that open new controller }

方法，您可以直接推送/模态/添加查看您想要的页面。

在“didFinishLaunchingWithOptions：”内部检查远程通知有效负载

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions {
 if (launchOptions[UIApplicationLaunchOptionsRemoteNotificationKey]) {
        NSDictionary *dict = launchOptions[UIApplicationLaunchOptionsRemoteNotificationKey];
        //Call your page here
    }
}

如何在点击应用图标时重定向到特定页面？

2 个答案: