Launch App using URL, but OpenUrl Not Called

Question

I have implemented a URL Scheme and use it to pass data to my app by calling method. The entire code is shown as below

- (BOOL)application:(UIApplication *)application handleOpenURL:(NSURL *)url{
    // Check the calling application Bundle ID
    if ([[url scheme] isEqualToString:@"yuvitime"])
    {
        NSLog(@"URL scheme:%@", [url scheme]);
        NSString * yuvitimeRequestValue = [url query];
        NSDictionary * userInfor = [[NSDictionary alloc]initWithObjectsAndKeys:yuvitimeRequestValue, @"YuvitimeRequest", nil];
        NSNotificationCenter * notificationCentre = [NSNotificationCenter defaultCenter];
        [notificationCentre postNotificationName:@"URLSCHEMEACTIVATEDNOTIFICATION" object:self userInfo:userInfor];

        return YES;
    }
    else
        return NO;
}

If my app is in the background, everything works fine. When you click a URL, the app is brought back to Foreground and the URL is handled as coded in the above function.

However, if the app is terminated (app not launched yet), by clicking the URL, it only launches the app without calling the handling function that is shown above.

After searching through, the best result i manage to get is this

application:WillFinishLaunchingWithOptions: When asked to open a URL, the return result from this method is combined with the return result from the application:didFinishLaunchingWithOptions: method to determine if a URL should be handled. If either method returns NO, the system does not call the application:openURL:options: method. If you do not implement one of the methods, only the return value of the implemented method is considered.

- application:didFinishLaunchingWithOptions: This method represents your last chance to process any keys in the launchOptions dictionary. If you did not evaluate the keys in your application:willFinishLaunchingWithOptions: method, you should look at them in this method and provide an appropriate response. Objects that are not the app delegate can access the same launchOptions dictionary values by observing the notification named UIApplicationDidFinishLaunchingNotification and accessing the notification’s userInfo dictionary. That notification is sent shortly after this method returns. The return result from this method is combined with the return result from the application:willFinishLaunchingWithOptions: method to determine if a URL should be handled. If either method returns NO, the URL is not handled. If you do not implement one of the methods, only the return value of the implemented method is considered.

Despite the explanation, i still do not know how to do it and i couldn't find anything else concrete online.

Thanks

Regards

Answer 1

I agree with Kaloyan, "handleOpenURL" is never called at application launch. So you have to check for URL in "launchOptions" in didFinishLaunchingWithOptions.

HOWEVER

I adopted the same solution as Apple example code for QuickActions (3D Touch). I keep the URL at launch in a variable, and I handle it in applicationDidBecomeActive:.

@interface MyAppDelegate ()
@property (nonatomic, strong) NSURL *launchedURL;
@end

@implementation MyAppDelegate

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{
    self.launchedURL = [launchOptions objectForKey:UIApplicationLaunchOptionsURLKey];
    ...
}

- (void)applicationDidBecomeActive:(UIApplication *)application
{
    if (self.launchedURL) {
        [self openLink:self.launchedURL];
        self.launchedURL = nil;
    }
}

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

    if (!openUrl)
    {
        return NO;
    }
    return [self openLink:openUrl];
}

- (BOOL)openLink:(NSURL *)urlLink
{
    ...
}

@end