Launcher.Default.TryOpenAsync() not working in MAUI App

Question

When calling Launcher.Default.TryOpenAsync(), its always return false in MAUI app. We need to open an installed app in our device from a button click. If not installed, it should go tor play store for download.

Tried this code, but not working

private async void Button1_ClickAsync(object sender, EventArgs e)
{
try
{   string package_name = txtPackagename.Text;   

    // bool supportsUri = await Launcher.Default.CanOpenAsync("com.google.android.youtube") // not working...

    // Try to open You tube App installed.
    bool launcherOpened = await Launcher.Default.TryOpenAsync("com.google.android.youtube");

    if (launcherOpened)
    {
        await DisplayAlert("Done", "Application Successfully Open.", "OK");
    }
}
catch (Exception ex)
{
    await DisplayAlert("No Support", ex.Message, "OK");
} 
}

Always its return false, even if the app is installed. How we can achieve this in MAUI? In Android studio, we can open an app with a package name.

Android Studio Code(this code working in Android Studio Project)

public void openApplication(Context context, String packageN) {
Intent i = context.getPackageManager().getLaunchIntentForPackage(packageN);
if (i != null) {
    i.addCategory(Intent.CATEGORY_LAUNCHER);
    context.startActivity(i);
} else {
    try {
        context.startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("market://details?id=" + packageN)));
    }
    catch (android.content.ActivityNotFoundException anfe) {
        context.startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("http://play.google.com/store/apps/details?id=" + packageN)));
    }
} }

Answer 1

Suppose the package name of the app you want to open is com.xamarin.secondapp.

Then you need to add IntentFilter and Exported =true to MainActivity.cs of the app you want to open(com.xamarin.secondapp).

Just as follows:

  [Activity(Label = "SecondApp", Icon = "@mipmap/icon", Theme = "@style/MainTheme", MainLauncher = true,Exported =true, ConfigurationChanges = ConfigChanges.ScreenSize | ConfigChanges.Orientation)] 
    [
    IntentFilter
    (
        new[] { Android.Content.Intent.ActionView },
        Categories = new[]
            {
                Android.Content.Intent.CategoryDefault,
                Android.Content.Intent.CategoryBrowsable
            },
        DataSchemes = new[] { "myapp" }
    )
]
    public class MainActivity : global::Xamarin.Forms.Platform.Android.FormsAppCompatActivity
    {
        protected override void OnCreate(Bundle savedInstanceState)
        {
            base.OnCreate(savedInstanceState);

            global::Xamarin.Forms.Forms.Init(this, savedInstanceState);
            LoadApplication(new App());
        }
    }

Note: remember to add code DataSchemes = new[] { "myapp" }.

For the current app, you need to add queries tag for the app you want to open in file manifest.xml.

For example:

  <?xml version="1.0" encoding="utf-8"?> 
<manifest xmlns:android="http://schemas.android.com/apk/res/android" android:versionCode="1" android:versionName="1.0" package="com.companyname.openappapp1">
    <uses-sdk android:minSdkVersion="21" android:targetSdkVersion="33" />
    <application android:label="OpenappApp1.Android" android:theme="@style/MainTheme">

      </application>
      <uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />


      <queries>
            <package android:name="com.xamarin.secondapp" />
      </queries>
</manifest>

And usage example:

public async void OpenSecondApp()
{

    var supportsUri = await Launcher.Default.CanOpenAsync("myapp://");
    if (supportsUri)
        await Launcher.Default.OpenAsync("myapp://com.xamarin.secondapp");

}