问题
I am writing code to launch other applications from my react native application for android and ios.
Using Linking form react native I am able to redirect to Play Store/App Store but
How can I launch App if it's already installed?
* I am getting the list of the app's from server
Linking.openURL('https://play.google.com/store/apps/details?id=com.example.myapp&hl=en')
Is there any way that I can launch the app if it's installed else redirect to App store/play store with respect to the platform?
Reference:react-native-app-link
回答1:
After many searches I have found an alternative for android without deep link URL is to use the native module react-native-intent-launcher to launch another app using package-name.
You can call the native function startActivity in react-native to do something with Intent which can only be solved with android native code.
Looking for iOS solution without deep link URL if any lead please update here
Once I found I will update Here
Thank you
回答2:
Your other app need to handle Deeplinking. If that's not already the case, have a look here for Android and here for iOS.
This will allow you to have your app's own URL scheme, for example testapp://example
Then you can simply use the Linking API, but instead of opening a HTTP URL, you can use you Deeplink URL scheme defined previously.
Linking.openURL('testapp://example');
来源：https://stackoverflow.com/questions/58856557/how-to-open-other-application-from-my-application