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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T15:54:25+00:00 2026-05-27T15:54:25+00:00

I want my Android app to appear listed as an option when the user

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I want my Android app to appear listed as an option when the user shares an URL from another app (like the browser). How can I register my app to do that? How can I react to link shares?

Thanks a lot.

Edit:

I’ve tried using IntentFilter like this without success:

<intent-filter>
    <action android:name="android.intent.action.VIEW" />
    <category android:name="android.intent.category.DEFAULT" />
    <category android:name="android.intent.category.BROWSABLE" />
</intent-filter>

Any ideas?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T15:54:26+00:00Added an answer on May 27, 2026 at 3:54 pm

    At the very bare minimum you need:

    <activity android:name=".ShareActivity">
        <intent-filter
            android:label="Share with my app">
            <action android:name="android.intent.action.SEND" />
            <category android:name="android.intent.category.DEFAULT" />
        </intent-filter>
    </activity>
    

    in your manifest…which will at least make it show up in the ‘share’ listing.

    The most important part you are missing is:

    <action android:name="android.intent.action.SEND" />
    

    To make it actually do something, you need an Activity.

    This may help too:
    http://sudarmuthu.com/blog/sharing-content-in-android-using-action_send-intent

    Additional Info:

    <activity android:name=".ShareActivity">
    <intent-filter
        android:label="Share with my app">
        <action android:name="android.intent.action.SEND" />
        <category android:name="android.intent.category.DEFAULT" />
        <data android:mimeType="text/plain" />
    </intent-filter>
    </activity>
    

    There the <data android:mimeType will limit what you respond to, if you want to limit your app’s response.

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