Android share url question for got chess

I want in android to add an option to share link. When I keep my finger pressed in a URL I want in the share options to appear the option "open PGN". I want to send this link in a new tab of my android web Brower with a prefix. Eg openpgn.com?Url=[the selected url]. How can I do that? Is there an easier way than to make a new android app from scratch?



_______

To achieve the functionality you're describing without starting a new Android app from scratch, you can create a simple Android application that registers itself as a share option for URLs. Below are the steps you need to follow:

### 1. Create a New Android Project

If you don't already have an Android project, you can create one using Android Studio:

1. Open Android Studio.
2. Click on `New Project`.
3. Choose `Empty Activity` and follow the prompts to set it up.

### 2. Modify the Android Manifest

Update the `AndroidManifest.xml` to declare your app as a sharer of URLs:

```xml
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.example.yourapp">

    <application
        ... >

        <activity android:name=".MainActivity">
            <intent-filter>
                <action android:name="android.intent.action.SEND" />
                <category android:name="android.intent.category.DEFAULT" />
                <data android:mimeType="text/plain" />
            </intent-filter>
        </activity>

    </application>
</manifest>
```

### 3. Handling the Shared URL

In your `MainActivity.java`, handle the intent in the `onCreate` method:

```java
import android.content.Intent;
import android.net.Uri;
import android.os.Bundle;
import android.widget.Toast;
import androidx.annotation.Nullable;
import androidx.appcompat.app.AppCompatActivity;

public class MainActivity extends AppCompatActivity {

    @Override
    protected void onCreate(@Nullable Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        // Handle the received intent
        Intent intent = getIntent();
        String action = intent.getAction();
        String type = intent.getType();

        if (Intent.ACTION_SEND.equals(action) && type != null) {
            if ("text/plain".equals(type)) {
                handleSendText(intent); // Handle text being sent
            }
        }
    }

    void handleSendText(Intent intent) {
        String sharedText = intent.getStringExtra(Intent.EXTRA_TEXT);
        if (sharedText != null) {
            // Here you can modify the URL and open it in a new tab
            String modifiedUrl = "https://openpgn.com?Url=" + Uri.encode(sharedText);
            Intent browserIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(modifiedUrl));
            startActivity(browserIntent);
        } else {
            Toast.makeText(this, "No URL shared", Toast.LENGTH_SHORT).show();
        }
    }
}
```

### 4. Test Your Application

1. Build and run your application on a real device or emulator.
2. Copy a URL from your browser or any app.
3. Long-press on the URL and select "Share".
4. Your application should appear in the share options. Select it, and it will open the modified URL in the browser.

### Note:

- Make sure your app handles multiple intents (like text, images, etc.) if necessary.
- Ensure that you have Internet permissions in your manifest if you're planning to work with a web URL (though not strictly necessary for opening links).
- If you want more customization or functionality, consider digging deeper into Android's share intents and the ways to interact with other apps.

### Conclusion

This is a relatively simple way to create functionality that allows users to share links to your application, which then redirects them to your specific URL with the desired prefix without building a full-fledged application from scratch.