CODEWITHSUNDEEP
iosflutterdarturl-launcherandroid-app-links
Amy
Amy

Reputation: 1390

How to open URL in external browser, bypassing app links?

I'm making a WebView flutter app and I've just added app links. But when I receive app links going to my websites homepages (e.g. 1st.money/home), I don't want those to open inside of the app, so I try and launch them back into the browser.

Here's my code so far:

import 'package:url_launcher/url_launcher.dart';
....
Future<void> openUrlInBrowser(String urlStr) async {
  final Uri url = Uri.parse(urlStr);
  if (!await launchUrl(
    url,
    mode: LaunchMode.externalApplication,
  )) {
    throw Exception('Could not launch $url');
  }
}

This opens the URL in the browser, but then the browser immediately sends it back to the app because it's registered as an app link. This creates a loop where the URL bounces between the app and the browser.

I've also tried:

But nothing has worked yet. Has anyone else also had the problem with this? Any help would be appreciated. Thank you :)

Upvotes: 0

Views: 160

Answers (1)

Shakh Rashidov
Shakh Rashidov

Reputation: 1

Step 1: Add url_launcher Dependency

Add the url_launcher package to your pubspec.yaml:


dependencies:
  flutter:
    sdk: flutter
  url_launcher: ^6.1.9

Step 2: Import and Use url_launcher In your Dart code, import the url_launcher package and use the launchUrl function to open the URL in an external browser.

import 'package:flutter/material.dart';
import 'package:url_launcher/url_launcher.dart';

void main() {
  runApp(MyApp());
}

class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return MaterialApp(
      home: Scaffold(
        appBar: AppBar(title: Text('Open URL in Browser')),
        body: Center(
          child: ElevatedButton(
            onPressed: () => _launchURL('https://www.example.com'),
            child: Text('Open URL'),
          ),
        ),
      ),
    );
  }

  Future<void> _launchURL(String url) async {
    final Uri uri = Uri.parse(url);
    if (await canLaunchUrl(uri)) {
      await launchUrl(
        uri,
        mode: LaunchMode.externalApplication,
      );
    } else {
      throw 'Could not launch $url';
    }
  }
}

Create a method to launch the URL: The _launchURL method checks if the URL can be launched, and if so, it uses launchUrl with LaunchMode.externalApplication to open the URL in the external browser. Using the method: When the button is pressed, the _launchURL method is called with the desired URL.

Upvotes: 0

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