Open external App from React native App ( Button Click)

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?

I think good is fine . Did you make a build after implement this code . — anil sidhu
– anil sidhu, Commented Aug 10, 2018 at 10:07
Yes i have done the build but It gives me error (no activity found to handle intent act=android.intent.action.view in react native) — Madhur
– Madhur, Commented Aug 10, 2018 at 10:08
I Hope you have react-native apps. from A app you want to open B app. right In a app use this code Linking.openURL('Test://app'); and in B app change Code : Android.manifestfile.xml file. right Please clear .. is i am getting right ? — anil sidhu
– anil sidhu, Commented Aug 10, 2018 at 10:13

anil sidhu · Accepted Answer · 2018-08-10 11:09:19Z

6

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)

answered Aug 10, 2018 at 11:09

anil sidhu

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8 Comments

Madhur Over a year ago

Sorry for late replying Anil. Nope its not working, I have applied above code simultaneously. When i clicked on buttoin its reflect and come back to the same place. ( like swipe to the right and same activity has open after that)...

Madhur Over a year ago

I have done above code in App B androidmanifesfile.xml in which android:host name would be same as my app folder name (testAppA) and now clicking from testAppA but it won't redirect to that ( testAppB ) Please help click is working but it won't open new react native application .

Madhur Over a year ago

If i am using this way of linking URL (Linking.openURL('testAppA://app')) it gives me error for not handling View intent

anil sidhu Over a year ago

can you show me the code of App B androidmanifesfile.xml and link code of a file please

Madhur Over a year ago

Code : folder name (AppFighter) i have written this code on ChrisApp manifest file. <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" /> <data android:scheme="app" android:host="appfighter" /> </intent-filter> Link to open Chris App (<Button title="Click me" onPress={ ()=>{ Linking.openURL('app://appfighter')}} />)

|

hitendra Savkare · Accepted Answer · 2018-08-10 11:10:59Z

0

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')
  }
});

answered Aug 10, 2018 at 11:10

hitendra Savkare

475 bronze badges

1 Comment

DevAS Over a year ago

it's doesn't work with Google Maps App other apps it's work very well, any idea?

Alien · Accepted Answer · 2018-08-13 09:29:35Z

-1

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

Here is the link you can check : https://snack.expo.io/rJm_YkqyW

edited Aug 13, 2018 at 9:29

Alien

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answered Aug 10, 2018 at 9:51

Arush Singh

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1 Comment

Madhur Over a year ago

This link is useful for browser URL, I want to open another App ( React Native) by clicking on the button

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