How do I open a web browser (URL) from my Flutter code?

If you target sdk 30 or above canLaunch will return false by default due to package visibility changes: https://developer.android.com/training/basics/intents/package-visibility

in the androidManifest.xml you'll need to add the following directly under <manifest>:

<queries>
    <intent>
        <action android:name="android.intent.action.VIEW" />
        <category android:name="android.intent.category.BROWSABLE" />
        <data android:scheme="https" />
    </intent>
</queries>

Then the following should word - for flutter 3 upwards:

const uri = Uri.parse("https://flutter.io");
if (await canLaunchUrl(uri)){
    await launchUrl(uri);
} else {
    // can't launch url
}

or for older versions of flutter use this instead:

const url = "https://flutter.io";
if (await canLaunch(url)){
    await launch(url);
} else {
    // can't launch url
}

launchUrl has a mode parameter which can be used to control where the url gets launched.

So, passing in launchUrl(uri, mode: LaunchMode.platformDefault) leaves the decision of how to launch the URL to the platform implementation. But you can also specify

LaunchMode.inAppWebView which will use an in-app web view LaunchMode.externalApplication for it to be handled by an external application

Or LaunchMode.externalNonBrowserApplication to be handled by a non-browser application.

TL;DR

This is now implemented as Plugin

const url = "https://flutter.io";
if (await canLaunchUrl(url))
  await launchUrl(url);
else 
  // can't launch url, there is some error
  throw "Could not launch $url";

---

Full example:

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

void main() {
  runApp(new Scaffold(
    body: new Center(
      child: new RaisedButton(
        onPressed: _launchURL,
        child: new Text('Show Flutter homepage'),
      ),
    ),
  ));
}

_launchURL() async {
  const url = 'https://flutter.io';
  if (await canLaunch(url)) {
    await launch(url);
  } else {
    throw 'Could not launch $url';
  }
}

In pubspec.yaml

dependencies:
  url_launcher: ^5.7.10

Special Characters:

If the url value contains spaces or other values that are now allowed in URLs, use

Uri.encodeFull(urlString) or Uri.encodeComponent(urlString) and pass the resulting value instead.