android - React Native - 如果应用程序 Y 已安装，应用程序 X 如何打开应用程序 Y

Question

如果安装了应用程序 Y，我希望我的应用程序(比如应用程序 X)想要打开应用程序 Y。并在 Android 和 iOS 中工作(如果未安装应用程序 Y，请打开 App Store/Play 商店)

我做了什么:

Linking.openURL(applicationStoreURL);

The code above success open Play Store/App Store of Application Y, but actually the application Y has been installed so i want to direct open to Application Y from Application X.

Answer 1

试试这个库

"Easily deep link to other apps in React Native. If the app isn't installed on the user's phone, open the App Store or Play Store link instead."

https://github.com/FiberJW/react-native-app-link

npm i -S react-native-app-link


import AppLink from 'react-native-app-link';

AppLink.maybeOpenURL(url, { appName, appStoreId, appStoreLocale, playStoreId }).then(() => {
  // do stuff
})
.catch((err) => {
  // handle error
});

AppLink.openInStore({ appName, appStoreId, appStoreLocale, playStoreId }).then(() => {
  // do stuff
})
.catch((err) => {
  // handle error
});