当APNS在iOS中收到打开视图控制器 [英] Open view controller when APNS is received in iOS

问题描述

嘿，我是新来的iPhone和我一直在尝试使用苹果推送通知。基本上，我想要做的是，当用户点击接收到推送通知的消息，然后我需要打开一个特定的视图控制器。我已经添加自定义数据与关键参数的类型我的有效载荷JSON，所以代通知类型的价值，我需要，而不是打开主视图控制器的特定视图控制器。

这是我的有效载荷JSON：

  {APS：{警告：这是测试消息，类型：通知，徽章：1，声音：默认 }}
 

我的code是：

   - （BOOL）应用：（*的UIApplication）的应用didFinishLaunchingWithOptions：（NSDictionary的*）launchOptions
{
    [UIApplication的sharedApplication] registerForRemoteNotificationTypes：
     （UIRemoteNotificationTypeBadge | UIRemoteNotificationTypeSound | UIRemoteNotificationTypeAlert）];    self.window = [[一个UIWindow页头] initWithFrame：方法[UIScreen mainScreen]界限]];    splashViewController = [[SplashViewController页头] initWithNibName：@SplashViewController捆绑：无];
        self.window.rootViewController = splashViewController;    [self.window makeKeyAndVisible]    返回YES;
} - （无效）应用：（*的UIApplication）的应用didReceiveRemoteNotification：（NSDictionary的*）USERINFO
{
    //这将调用时，你的应用将会从服务器得到任何通知。    [UIApplication的sharedApplication] setApplicationIconBadgeNumber：0];
    [UIApplication的sharedApplication] cancelAllLocalNotifications]    *的NSDictionary APS =（的NSDictionary *）[USERINFO objectForKey：@APS];
    *的NSMutableString = notificationType [APS objectForKey：@型];    的NSLog（@通知类型是=％@，notificationType）;    //对于重定向到视图
    *的UIViewController的viewController;
    如果（[notificationType isEqualToString：@通知]）{        //更新通知
        UpdatesViewController * uvc1 = [[UpdatesViewController页头] initWithNibName：@UpdatesViewController捆绑：无];
        的viewController = uvc1;
    }
    其他{
        //投票和放大器; QA
        VotingViewController * votingViewController = [[VotingViewController页头] initWithNibName：@VotingViewController捆绑：无];
         的viewController = votingViewController;
    }     [self.window.rootViewController presentViewController：的viewController动画：是完成：NULL];
}
 

我的code不是presenting其他视图控制器。正如我在code所提到的，我得到了两种Web服务（通知类型）的通知1和2苹果表决。我不知道如何代表通知类型的present特定视图控制器，如果类型是通知，那么我需要开UpdatesViewController否则它将打开其他的ViewController。如果有人知道如何做到这一点，请帮助我。

感谢。

解决方案

  //对于重定向到视图
    *的UIViewController UVC;
    如果（[notificationType isEqualToString：@通知]）{        *的UIViewController更新= [[UpdatesViewController页头] initWithNibName：@UpdatesViewController捆绑：无];
        //添加更新的具体数据updates.data = [APS objectForKey：@数据];
        UVC =更新;
    }    否则，如果（[notificationType isEqualToString：@投票]）{
        *的UIViewController投票= [[VotingViewController页头] initWithNibName：@VotingViewController捆绑：无];
        //添加投票的具体数据voting.data = [APS objectForKey：@数据];
        UVC =投票;
    }
    [self.window.rootViewController presentViewController：UVC动画：是完成：NULL];
 

Hey I'm new to iPhone and I have been trying to use an Apple push notification. Basically, what I want to do is that when user clicks on the received push notification message, then i need to open a specific view controller. I have added custom data with key parameter "type" to my payload JSON, so on behalf of notification type value i need to open particular view controller instead of the main view controller.

here is my payload JSON :

{"aps":{"alert":"This is testing message","type":"Notify","badge":1,"sound":"default"}}

my code is :

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{
    [[UIApplication sharedApplication] registerForRemoteNotificationTypes:
     (UIRemoteNotificationTypeBadge | UIRemoteNotificationTypeSound | UIRemoteNotificationTypeAlert)];

    self.window = [[UIWindow alloc] initWithFrame:[[UIScreen mainScreen] bounds]];

    splashViewController = [[SplashViewController alloc] initWithNibName:@"SplashViewController" bundle:nil];
        self.window.rootViewController = splashViewController;

    [self.window makeKeyAndVisible];

    return YES;
}

- (void)application:(UIApplication*)application didReceiveRemoteNotification:(NSDictionary*)userInfo
{
    //this will call when your app will get any notification from server.

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

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

    NSLog(@"notification type is = %@", notificationType); 

    //For redirect to the view
    UIViewController *viewController;
    if([notificationType isEqualToString:@"Notify"]){

        //Notify updates
        UpdatesViewController *uvc1 = [[UpdatesViewController alloc] initWithNibName:@"UpdatesViewController" bundle:nil];
        viewController = uvc1;
    }
    else {
        //Voting & QA
        VotingViewController *votingViewController = [[VotingViewController alloc] initWithNibName:@"VotingViewController" bundle:nil];
         viewController = votingViewController;
    }

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

My code is not presenting other view controller. As mentioned in my code, i am getting two kind of web service (notification type) 1. Notify and 2. Voting from Apple. i dont know how to present specific view controller on behalf of notification type, if type is "notify" then i need open UpdatesViewController else it will open other viewcontroller. If anyone knows how to do this, please help me.

Thanks.

解决方案

//For redirect to the view
    UIViewController *uvc;
    if([notificationType isEqualToString:@"Notify"]){

        UIViewController *updates = [[UpdatesViewController alloc] initWithNibName:@"UpdatesViewController" bundle:nil];
        // add updates specific data updates.data = [aps objectForKey:@"data"];
        uvc = updates;
    }

    else if([notificationType isEqualToString:@"Voting "]) {
        UIViewController *voting = [[VotingViewController alloc] initWithNibName:@"VotingViewController" bundle:nil];
        // add voting specific data voting.data = [aps objectForKey:@"data"];
        uvc = voting;
    }
    [self.window.rootViewController presentViewController:uvc animated:YES completion:NULL];

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