NSWidgetExtensionContext openURL Swift

Question

I have been attempting to implement a button to open my iOS app from its widget. I realize this issue has been beaten to death on the forums but I cannot find explanation with the specific error I am receiving. Perhaps some of you more experienced iOS developers can shed some light on this.

I am developing an update to one of my iOS apps for iOS 10 using XCode 8.1 and Swift 2.

Code for my widget's button:

URL scheme added to the widget's info.plist:

The runtime error I receive when pressing the OpenApp button:

AppWidget[11872:3577323] __55-[_NCWidgetExtensionContext openURL:completionHandler:]_block_invoke failed: Error Domain=NSOSStatusErrorDomain Code=-10814 "(null)"

// Note: app name has been substituted with appropriate generics.

Answer 1

I often find the OS Status lookup site pretty useful to infer details from errors. An OS error with code -10814 is a kLSApplicationNotFoundErr, which describes the scenario when:

No application in the Launch Services database matches the input criteria.

It sounds like your application has not been properly registered with the system as a consumer of the URL scheme you are using. Have you double-double (double!) checked that the bundle identifier and URL scheme match? Have you verified that your app can be launched with the URL from Safari?