How to open the app when someone clicks something on an Apple Watch app?

Question

I am making an Apple Watch app. One of the buttons will open the iphone app connected to the watch app.

What code do I use to do this?

I don't know what to even try?

Note: I am using swift for this project.

score 1 · Answer 1 · answered May 07 '15 at 11:39

1

WatchKit doesn't include the ability to open the host iOS app in the foreground. The best you can do is open it in the background using openParentApplication:reply:.

If you need the user to do something in your iOS app, consider making use of Handoff.

answered May 07 '15 at 11:39

bgilham

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Whats a handoff. Please elaborate? – nachshon fertel May 07 '15 at 11:41
@nachshonfertel some info about handoff here; https://developer.apple.com/library/ios/documentation/UserExperience/Conceptual/Handoff/HandoffFundamentals/HandoffFundamentals.html You should searched this for yourself. – Bas May 07 '15 at 11:42
Is there any way to launch a call on the iphone? – nachshon fertel May 07 '15 at 11:42
@nachshonfertel search for it on the internet. Show some effort. – Bas May 07 '15 at 11:43
I did, but i got no answer. Please help. – nachshon fertel May 07 '15 at 11:44
4

@nachshonfertel There's a link here in the comments start there. SO isn't here to hold your hand every step of the way. – bgilham May 07 '15 at 11:46

score 0 · Answer 2 · edited May 23 '17 at 11:53

0

It is not possible to activate an inactive iPhone app from the Watch. It is, however, possible to call the iPhone app to perform a task or to ask for data. See here: Calling parent application from Watch app

edited May 23 '17 at 11:53

Community

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answered May 07 '15 at 11:40

John

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score 0 · Answer 3 · answered May 21 '15 at 11:19

You can only open the iPhone app in the background by the following method:

Swift:

openParentApplication([ParentApp], reply:[Reply])

Objective-C:

openParentApplication:reply:

There is no ability to open the parent app in the foreground.

Note: To send data to iOS app in the background, use the first method.

Note: According to bgilham,

If you need the user to do something in your iOS app, consider making use of Handoff.

score 0 · Answer 4 · answered May 27 '15 at 11:17

If you need to open your parent app in the foreground, use Handoff!

https://developer.apple.com/handoff/

Example:

Somewhere shared for both:

static let sharedUserActivityType = "com.yourcompany.yourapp.youraction"
static let sharedIdentifierKey = "identifier"

on your Watch:

updateUserActivity(sharedUserActivityType, userInfo: [sharedIdentifierKey : 123456], webpageURL: nil)

on your iPhone in App Delegate:

func application(application: UIApplication, willContinueUserActivityWithType userActivityType: String) -> Bool {
    if (userActivityType == sharedUserActivityType) {
        return true
    }
    return false
}

func application(application: UIApplication, continueUserActivity userActivity: NSUserActivity, restorationHandler: ([AnyObject]!) -> Void) -> Bool {
    if (userActivity.activityType == sharedUserActivityType) {
        if let userInfo = userActivity.userInfo as? [String : AnyObject] {
            if let identifier = userInfo[sharedIdentifierKey] as? Int {
                //Do something
                let alert = UIAlertView(title: "Handoff", message: "Handoff has been triggered for identifier \(identifier)" , delegate: nil, cancelButtonTitle: "Thanks for the info!")
                alert.show()
                return true
            }
        }
    }
    return false
}

And finally (this step is important!!!): In your Info.plist(s)

enter image description here

How to open the app when someone clicks something on an Apple Watch app?

4 Answers4