Andorid，如何使用自定义URL方案打开另一个应用程序？

Question

From my native application, I need to open my another app which build by Ionic.

I tried like this,

val intent = packageManager.getLaunchIntentForPackage("com.example.app")
intent.putExtra("uri", "hello 2?")
intent.data = Uri.parse("hello?")
startActivity(intent)

It could launch the app but, it never called handleOpenURL() method.

So I want to try using the custom URL scheme instead of the package name.

How can I open another app using custom URL scheme with URI parameter?

Answer 1

 Intent intent = new Intent(Intent.ACTION_VIEW);
    intent.setData(Uri.parse("YOUR_URL"));
    intent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
    if (intent.resolveActivity(getPackageManager()) != null) {
        startActivity(intent);
    } else {
        Toast.makeText(this, R.string.no_activity_found, Toast.LENGTH_LONG).show();
    }

Answer 2

This worked for me. In your android manifest's launcher activity mention the following lines

<intent-filter>
      <data android:scheme="http" />
      <!-- or you can use deep linking like  -->

      <data android:scheme="http" android:host="abc.xyz"/>
      <action android:name="android.intent.action.VIEW"/>
      <category android:name="android.intent.category.BROWSABLE"/>
      <category android:name="android.intent.category.DEFAULT"/>

</intent-filter>

Instead of this host name change to suit your preferences and you can use your own scheme as well.