android：React native 从另一个应用程序打开一个应用程序？

Question

I am trying to open a another app( https://play.google.com/store/apps/details?id=com.inova.velocity ) from my app. But there are all the tutorial just redirecting url to playstore only.(I found a github link( https://github.com/FiberJW/react-native-app-link ) and it opens the app for iOS only, but for Android it is redirecting to playstore). Is there is any way to solve this problem?

Linking.canOpenURL('market://details?id=com.inova.velocity')
      .then((canOpen) => {
        if (canOpen) { 
          console.log('open app'); 
          return Linking.openURL('market://details?id=com.inova.velocity')
                 };
        }).catch(err => console.log('An error occurred', err));

Answer 1

Yes your code is correct. But you have used playstore url, instead of schema url. you have to set the schemaUrl which you can get it from relevant app developer. if there is no schema url set for the that app you can't open it. after you get the SchemaUrl you can use your code. like below.

Linking.canOpenURL(SchemaUrl).then(supported => {
             if (supported) {
               console.log('accepted');
               return Linking.openURL(SchemaUrl);
             } else {
               console.log('an error occured');
             }
           }).catch(
             err => console.log('an error occured');
           );

Answer 2

100% working for open all app from another app using react-native-send-intent .

React Native Android module to use Android's Intent actions for opening third party apps.

Installation npm install react-native-send-intent --save

Register Module >= 0.29 (in MainApplication.java) Adding only 2 lines

import com.burnweb.rnsendintent.RNSendIntentPackage;  // <--- import in MainApllication.java

public class MainApplication extends Application implements ReactApplication {
  ......

  @Override
  protected List<ReactPackage> getPackages() {
  return Arrays.<ReactPackage>asList(
        new MainReactPackage(),
        new RNSendIntentPackage()); // <------ add this line to your MainApplication 
  class
  }

  ......

  }

Example / Open App in your react-native code

 SendIntentAndroid.isAppInstalled('com.medlife.customer').then((isInstalled) => {
            if (isInstalled) {
                SendIntentAndroid.openApp('com.medlife.customer').then((wasOpened) => {
                });
                console.log("is installed true");
            }
            else {
                Linking.openURL('https://play.google.com/store/apps/details?id=com.medlife.customer&hl=en').catch(err => {
                    console.log(err)
                })
            }
        });

I am opening 3rd party Medlife app from my app if you have need to open another app then only change pacakage name in SendIntentAndroid.openApp('com.medlife.customer')

react-native-send-intent git hub example here

Answer 3

using react-native-send-intent module , you can do
SendIntentAndroid.openApp('packagename').then((wasOpened) => {}); where your package name is whatever application package name you want to open.

for eg SendIntentAndroid.openApp('com.inova.velocity').then((wasOpened) => {});

wasOpened is a Boolean promise telling you whether the app was opened or not