Intercepting links from the browser to open my Android app

Question

I'd like to be able to prompt my app to open a link when user clicks on an URL of a given pattern instead of allowing the browser to open it. This could be when the user is on a web page in the browser or in an email client or within a WebView in a freshly-minted app.

For example, click on a YouTube link from anywhere in the phone and you'll be given the chance to open the YouTube app.

How do I achieve this for my own app?

Answer 1

Use an android.intent.action.VIEW of category android.intent.category.BROWSABLE.

From Romain Guy's Photostream app's AndroidManifest.xml,

    <activity
        android:name=".PhotostreamActivity"
        android:label="@string/application_name">

        <!-- ... -->            

        <intent-filter>
            <action android:name="android.intent.action.VIEW" />
            <category android:name="android.intent.category.DEFAULT" />
            <category android:name="android.intent.category.BROWSABLE" />
            <data android:scheme="http"
                  android:host="flickr.com"
                  android:pathPrefix="/photos/" />
            <data android:scheme="http"
                  android:host="www.flickr.com"
                  android:pathPrefix="/photos/" />
        </intent-filter>
    </activity>

Once inside you're in the activity, you need to look for the action, and then do something with the URL you've been handed. The Intent.getData() method gives you a Uri.

    final Intent intent = getIntent();
    final String action = intent.getAction();

    if (Intent.ACTION_VIEW.equals(action)) {
        final List<String> segments = intent.getData().getPathSegments();
        if (segments.size() > 1) {
            mUsername = segments.get(1);
        }
    }

It should be noted, however, that this app is getting a little bit out of date (1.2), so you may find there are better ways of achieving this.