How to open Android app if installed, otherwise redirect to Google Play (no JavaScript)

I’d like to add a button to my Android app that acts as a link to another app. If the user has already downloaded it, the app launches; otherwise, it opens the Google Play Store with that app listed. I have access to both apps, and I’m wondering how I could set this up without using JavaScript redirects on the server.

0
Apr 30 at 10:18 AM
User AvatarStanisław Olszak
#android#android-deep-link

Accepted Answer

Open Android app if installed, otherwise redirect to Google Play

Required steps for target app:

  1. Add an intent filter for deep links:
<intent-filter android:autoVerify="true">
<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" />
    <data android:scheme="https" />
    <data android:host="example.com" />

</intent-filter>

  1. Handle the deep link in the navigation graph or MainActivity

  2. Configure the .well-known/assetlinks.json file on your server (see documentation).

Requirements for devices with a button:

  • Use a function to open the app with a fallback to Google Play:
fun openApp(context: Context) {
    val url = "https://example.com/" // your deep link
    val packageName = "com.app.example" // target app package name

    val intent = Intent(Intent.ACTION_VIEW, url.toUri()).apply {
        setPackage(packageName) // Only this app can handle the intent
        addFlags(Intent.FLAG_ACTIVITY_NEW_TASK)
    }

    try {
        context.startActivity(intent)
    } catch (e: ActivityNotFoundException) {
        // Google Play Store fallback
        val playIntent = Intent(
            Intent.ACTION_VIEW,
            "https://play.google.com/store/apps/details?id=$packageName".toUri()
        )
        context.startActivity(playIntent)
    }

}
User AvatarStanisław Olszak
Apr 30 at 10:18 AM
0