Want to open a URL like https://www.wizbrand.com/login directly in a mobile browser from your Flutter app?

What You Want:

When a user clicks a button in your Flutter app, the app should open the website in Chrome (Android), Safari (iOS), or another browser — outside of your app.

The Right Way to Do It (Step-by-Step)

1. Install url_launcher Package

Open your pubspec.yaml and add:

dependencies:
  url_launcher: ^6.2.5  # Or latest version

Then run:

flutter pub get

2. Import the Package

In the Dart file where you want to open the browser, import:

import 'package:url_launcher/url_launcher.dart';

3. Write the Function (THE RIGHT WAY)

Here’s the correct and safe way to launch a website in the browser:

Future<void> _launchWebsite() async {
  final Uri uri = Uri.parse('https://www.wizbrand.com/login');

  if (!await launchUrl(uri, mode: LaunchMode.externalApplication)) {
    if (context.mounted) {
      ScaffoldMessenger.of(context).showSnackBar(
        const SnackBar(content: Text('Could not launch website')),
      );
    }
  }
}

Why This Works:

• Uri.parse(...) ensures the URL is valid and safe.
• launchUrl(..., mode: LaunchMode.externalApplication) tells Flutter to open it in a browser, not inside your app.
• context.mounted avoids errors when showing SnackBar after the widget is disposed.
• SnackBar provides user feedback if something goes wrong.

4. Call It From a Button

ElevatedButton(
  onPressed: _launchWebsite,
  child: Text('Open Wizbrand Login'),
)

The WRONG Way (Common Mistakes to Avoid)

Mistake 1: Using launch Instead of launchUrl

// ❌ Deprecated or outdated
await launch('https://www.wizbrand.com'); 

👉 This used to work in older versions of url_launcher, but it’s now outdated.

Mistake 2: Not Using Uri.parse(...)

// ❌ Wrong: passing a String instead of Uri
await launchUrl('https://www.wizbrand.com'); 

Will give a type error — launchUrl() needs a Uri, not a plain string.

Mistake 3: Forgetting Platform Permissions

If you don’t do platform setup, it won’t open on Android/iOS.

Android:

In android/app/src/main/AndroidManifest.xml:

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

Also, ensure your minSdkVersion is at least 21 in build.gradle.

iOS:

In ios/Runner/Info.plist, add:

<key>LSApplicationQueriesSchemes</key>
<array>
  <string>https</string>
</array>
<key>NSAppTransportSecurity</key>
<dict>
  <key>NSAllowsArbitraryLoads</key>
  <true/>
</dict>

Bonus: Clickable Text URL

Want to make a text link clickable too?

InkWell(
  onTap: _launchWebsite,
  child: Text(
    'www.wizbrand.com/login',
    style: TextStyle(
      color: Colors.blue,
      decoration: TextDecoration.underline,
    ),
  ),
)

Final Thoughts

Opening URLs in the browser may seem small, but doing it the right way ensures:

• Users get a smooth experience
• Your app works on all platforms (Android, iOS, Web)
• You’re future-proof with Flutter’s latest standards