How can I open my Android app with a custom link?

Question

I Wanna open My android app by links like WhatsApp and Telegram.

(example) https://chat.whatsapp.com if I Click this link and WhatsApp is installed this link will open WhatsApp So how can I do that in my app?

Answer 1

If you want to deep link your app. For example: You open a link and it should be open through your app. In this case I use a website opened with a webView in your app. You can customize that of course.

Start to create an <intent-filter> in your AndroidManifest.xml :

      <application
         <activity
            <intent-filter>
                   ...
                <action android:name="android.intent.action.VIEW"></action>
                <category android:name="android.intent.category.DEFAULT"></category>
                <category android:name="android.intent.category.BROWSABLE"></category>
                   
                <data android:scheme="https"
                    android:host="yourURL"></data>
                   ...
            </intent-filter>
        </activity>
     </application>

And in your MainActivity.java you write the following code to get the data to be set in the webView :

Uri uri = getIntent().getData();
        if (uri!=null){
            String path = uri.toString();
            
            WebView webView = (WebView)findViewById(R.id.webView);
            webView.loadUrl(path);
        }

And your webView defined in your activity_main.xml :

                 <WebView
                    android:layout_width="match_parent"
                    android:layout_height="match_parent"
                    android:id="@+id/webView"/>

That's it. Good luck: Cheers :)

Answer 2

First thing you should define your base URL inside strings file like this:

<string name="base_url" translatable="false">yourapp.me</string>

after that you should define an IntentFilter inside Manifest file like this:

<activity android:name=".ExampleActivity">


<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="https" android:pathPrefix="/api" android:host="@string/base_url"/>

</intent-filter>

</activity>

So your link should be like this:

https://yourapp.me/api

when you click on this link it should open your app in the activity that you put this intent filter inside it.