Somehow not working to me. Anyone works?

[WEI-TING LIN]

I ran into one problem and couldn't figure it out what happened. I followed the instructions in the README and also checked the example code, however in performActionForShortcutItem function

func application(application: UIApplication, performActionForShortcutItem shortcutItem:     UIApplicationShortcutItem, completionHandler: Bool -> Void) {
        guard let quickActions = quickActions else { return completionHandler(false) }
        guard let rootViewController = window?.rootViewController else { return completionHandler(false) }
        completionHandler(quickActions.handle(rootViewController, shortcutItem: shortcutItem))
}

This line always return false

quickActions.handle(rootViewController, shortcutItem: shortcutItem)

I don't know if it is because I set something wrong or it is because of the library.

[Ricardo Pereira]

@Tim77277 You're using the latest version? Did you look the example? https://github.com/ricardopereira/QuickActions/tree/master/Example

[WEI-TING LIN]

UPDATE:
I just solved the problem. It wasn't library's problem. The reason it didn't work before was because I used a UISplitViewController in my project. So when the function inside completionHandler got called

quickActions.handle(rootViewController, shortcutItem: shortcutItem)

It always returned false simply because my rootViewController was not the controller that I want to handle the QuickActionSupport protocol.

Hopefully it would help some people who encounter the same problem as me.

[WEI-TING LIN]

UPDATE:
I just solved the problem. It wasn't library's problem. The reason it didn't work before was because I use a UISplitViewController in my project. So when the function inside completionHandler got called

quickActions.handle(rootViewController, shortcutItem: shortcutItem)

It always returned false simply because my rootViewController is not the controller that I want to handle the QuickActionSupport protocol.

Hopefully it would help some people who encounter the same problem as me.

[Ricardo Pereira]

Ok, thanks for sharing!