About starting Android app from an URL

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You need to have two <intent-filter> elements for this <activity>. One will be for MAIN and LAUNCHER. The other will be for VIEW, BROWSABLE/DEFAULT, and your <data> element:

<activity android:name=".MyActivity">
    <intent-filter>
        <action android:name="android.intent.action.MAIN"></action>
        <category android:name="android.intent.category.LAUNCHER"></category>
    </intent-filter>
    <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:host="my.app" android:scheme="http"></data>
    </intent-filter>
</activity>

Then, http://my.app should launch your activity.

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FelipeDev.-
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FelipeDev.-

Updated on June 04, 2022

Comments

  • FelipeDev.-
    FelipeDev.- about 2 years

    I am trying to figure out how to start an app from a URL, and how I should write that URL.

    I have the following code in my AndroidManifest:

    <activity android:name=".MyActivity">
        <intent-filter>
            <action android:name="android.intent.action.MAIN"></action>
            <action android:name="android.intent.action.VIEW"></action>
            <category android:name="android.intent.category.LAUNCHER"></category>
            <category android:name="android.intent.category.DEFAULT"></category>
            <category android:name="android.intent.category.BROWSABLE"></category>
            <data android:host="my.app" android:scheme="http"></data>
        </intent-filter>
    </activity>
    

    I used a URL as explained in this answer, but nothing happens.

    Please let me know if my intent is well written, and how I should write the URL that calls that app, and note that I need to call my "Main" Activity.