从 React Native App 打开外部应用程序（按钮单击）

Question

I want to open New React native App by clicking on Button in which I have used

Linking Concepts in React native

React native Code : Test is the name of the Other App

Linking.openURL('Test://app');

Also following URL for adding Intent in the android.manifest.xml file Andriod Intent

Code : Android.manifestfile.xml

  <activity
        android:name=".MainActivity"
        android:launchMode="singleTask"
        android:label="@string/app_name"
        android:configChanges="keyboard|keyboardHidden|orientation|screenSize"
        android:windowSoftInputMode="adjustResize">
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />
            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
          <intent-filter>
              <action android:name="android.intent.action.VIEW" />
              <category android:name="com.Test" />
              <category android:name="android.intent.category.BROWSABLE" />
              <data android:scheme="Test" android:host="app" />
          </intent-filter>
      </activity>

How can I resolve the issue?

Answer 1

Add this code in your AndroidManifest.xml file parallel to current intent filter

 <intent-filter android:label="@string/app_name">
        <action android:name="android.intent.action.VIEW" />
        <category android:name="android.intent.category.DEFAULT" />
        <category android:name="android.intent.category.BROWSABLE" />
        <!-- Accepts URIs that begin with "example://gizmos” -->
        <data android:scheme="app"
              android:host="testApp" />
    </intent-filter> 

and run this command React-native run-android

Add below code to your react-native file.

<Button title="Click me" onPress={ ()=>{ Linking.openURL('app://testApp')}} />

for save the time . You can to both code in same application and same application will be open on button press

i just try this code and its working for me let me know if still facing issue (Y)

Answer 2

You can open the external application by using following code

Linking.canOpenURL("fb://app").then(supported => {
  if (supported) {
    Linking.openURL("fb://app");
  } else {
    alert('sorry invalid url')
  }
});

Answer 3

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 4

<Button title="Click me" onPress={ ()=>{ Linking.openURL('https://google.com')}} />

这是您可以查看的链接： https : //snack.expo.io/rJm_YkqyW