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I have 2 iOS apps that have watchOS 2 extensions. If a user is using the first app, it sometimes makes sense for them to want to launch into the second app. In iOS you can do this using openURL:

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:customURL]];

But I can not find an equivalent on watchOS 2. Does anyone know of a way to have a watchOS 2 app launch another app?

lehn0058
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  • If you combined ideas from these two posts then wouldn't it help? – Sam B Dec 07 '15 at 16:45
  • http://stackoverflow.com/questions/6123194/how-to-use-uiapplication-sharedapplication-openurl-open-other-app – Sam B Dec 07 '15 at 16:46
  • http://stackoverflow.com/questions/27700907/launch-host-app-from-watch-app – Sam B Dec 07 '15 at 16:46
  • Unfortunately, those posts are about launching the Apple Watch's parent app. Those API's won't work to try to launch a different app on the watch. – lehn0058 Dec 07 '15 at 18:34

1 Answers1

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Unfortunately, You can't.

You can launch only system apps in watchOS using [WKExtension openSystemURL:]. (Some system URLs may works on only iOS)

There is no API for 3rd party scheme.

jeeeyul
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