如何以编程方式从iOS应用程序打开Apple Watch配套应用程序 [英] How to programmatically open Apple Watch companion app from iOS app

问题描述

我想向用户提示，我的iOS应用程序支持Apple Watch。

I would like to give a hint to the user, that my iOS app supports the Apple Watch.

所以我想从我的iOS应用程序中链接/打开Apple Watch配套应用程序，非常类似于使用（打开设置应用程序[[UIApplication sharedApplication] openURL：[NSURL URLWithString：UIApplicationOpenSettingsURLString]]; ）

So I would like to link/open the Apple Watch companion app from within my iOS app, very much similar to opening the Settings App using ([[UIApplication sharedApplication] openURL: [NSURL URLWithString:UIApplicationOpenSettingsURLString]];)

这将使用户能够直接导航到Watch伴侣应用程序为手表设置我的应用程序。

This shall enable the user to directly navigate to the Watch companion app to setup my app for the watch.

我找不到任何可以打开常用应用程序或特定于配套应用程序部分的URL。

I could not find any URL which would open the companion app in general or specific to a section of the companion app.

如果不支持直接链接，我也对此用例的替代方法感兴趣。

If direct linking is not supported, I am also interested in alternative approaches for this use case.

提前致谢！

编辑：
我检查了模拟器中的WatchKitSettings Info.plist文件，看它是否注册了任何URL架构，但是它没有。

I checked the WatchKitSettings Info.plist file in the simulator to see if it registers any URL schema, but it does not.

推荐答案

我想不可能以编程方式从iOS启动手表伴侣应用程序。相反的方式是可能的：在收到来自手表的消息后在后台启动iOS应用程序。请参阅WWDC演讲介绍Watch Connectivity 。

I guess it's not possible to programmatically launch the watch companion app from iOS. The opposite way would be possible: to launch the iOS app in the background upon receiving a message from the watch. See WWDC talk Introducing Watch Connectivity.

你可以检查 WCSession.defaultSession（）。watchAppInstalled 并要求用户启动应用程序，如果它是真的。

You could check WCSession.defaultSession().watchAppInstalled and ask the user to launch the app if it is true.

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