CODEWITHSUNDEEP
flutterdarturlbrowserwebbrowser-control
Seth Ladd
Seth Ladd

Reputation: 120439

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

I am building a Flutter app, and I'd like to open a URL into a web browser or browser window (in response to a button tap). How can I do this?

Upvotes: 228

Views: 350777

Answers (12)

user24746231
user24746231

Reputation: 1

click on the blue texts for images Make sure to make changes according to the url_launcher package as this may change in new versions:Change sdkVersion then make changes in your AndroidManisfest.xml file as below Add the action and data in then finally make changes for ios devices in the info.plist file as below Add permissions in info.plist then finally implement the code yow will see it redirected to the flutter website took mine 4 hours just because I did not place the intents correctly

this image has the source code [4]: https://i.sstatic.net/196vJug3.png

Sorry I had ho use pictures as I am new here but wanted to help and this stupid rules. Moreover you should keep eye on the imports too as the link is also available in "dart:\io" package avoid it and use import of the url launcher

Upvotes: 0

Sisalik
Sisalik

Reputation: 82

Since the question doesn't ask the link being a button, I have two solutions.

1. With the button

Many users here provide solutions that are similar, but for me they don't quite work for different reasons.

Use url_launcher and add it to the dependencies in pubspec.yaml.

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

void main() {
  runApp(new Scaffold(
    body: new Center(
      child: new ElevatedButton( //RaisedButton doesn't exist
        onPressed: () async { // URL launching is an async function, the button must wait for press
          _launchURL(Uri.https('google.com', ''));
        },
        child: new Text('Show Google homepage'),
      ),
    ),
  ));
}

Future<void> _launchURL(url) async { //you can also just use "void" or nothing at all - they all seem to work in this case
  if (!await launchUrl(url)) {
    throw Exception('Could not launch $url');
  }
}

Wrong

  • onPressed: _launchURL with void _launchURL() async { gives an error This expression has a type of 'void' so its value can't be used.
  • onPressed: _launchURL with Future<void> _launchURL() async { gives an error The argument type 'Future<void>' can't be assigned to the parameter type 'void Function()?'.
  • onPressed: _launchURL with _launchURL() async { launches automatically and throws an Exception type 'Future<dynamic>' is not a subtype of type '(() => void)?'
  • canLaunchUrl and canLaunchUrlString give messages I/UrlLauncher( 2899): component name for https://google.com is null and I/UrlLauncher( 2899): component name for https://flutter.dev is null and the app never launches your URL.

Short way for using this:

      ...
      child: new ElevatedButton(
        onPressed: () async {
          if (!await launchUrl(Uri.https('google.com', ''))) {
            throw Exception('Could not launch');
          }
        },
        ...

Using !await launchUrl checks if the URL can be launched and if not, then throws an Exception.

2. Without the button (just text, like a hyperlink)

Use url_launcher and add it to the dependencies in pubspec.yaml.

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

void main() {
  runApp(new Scaffold(
    body: new Center(
      child: new InkWell (
        onTap: () async {
          _launchURL(Uri.https('google.com', ''));
        },
        child: new Text('Show Google homepage'),
      ),
    ),
  ));
}

Future<void> _launchURL(url) async {
  if (!await launchUrl(url)) {
    throw Exception('Could not launch $url');
  }
}

Upvotes: 1

Hasan ko&#231;
Hasan ko&#231;

Reputation: 169

If you want to open it in a local browser, the url_launcher package has a feature like this:

launchUrlString('https://exemple.com,mode: LaunchMode.externalApplication);

Upvotes: 0

G&#252;nter Z&#246;chbauer
G&#252;nter Z&#246;chbauer

Reputation: 657018

TL;DR

This is now implemented as Plugin

 final Uri url = Uri.parse('https://flutter.dev');
 if (!await launchUrl(url)) {
      throw Exception('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 {
   final Uri url = Uri.parse('https://flutter.dev');
   if (!await launchUrl(url)) {
        throw Exception('Could not launch $_url');
    }
}

In pubspec.yaml

dependencies:
  url_launcher: ^6.1.11

Check out the latest url_launcher package.

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.

Upvotes: 352

Kab Agouda
Kab Agouda

Reputation: 7269

The best way is to use url_launcher package .

Add url_launcher as a dependency in your pubspec.yaml file.

dependencies:
  url_launcher:

An example of how to use it :

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

void main() {
  runApp(
    MaterialApp(
        home: Scaffold(
      appBar: AppBar(title: Text('Flutter is beautiful'),),
      body: Center(
        child: RaisedButton(
          onPressed: _launchURL,
          child: Text('Show Flutter homepage'),
        ),
      ),
    )),
  );
}

_launchURL() async {
  const url = 'https://flutter.dev';
  if (await canLaunchUrl(Uri.parse(url))) {
    await launchUrl(Uri.parse(url));
  } else {
    throw 'Could not launch $url';
  }
}

Output : enter image description here

The launch method takes a string argument containing a URL . By default, Android opens up a browser when handling URLs. You can pass forceWebView: true parameter to tell the plugin to open a WebView instead. If you do this for a URL of a page containing JavaScript, make sure to pass in enableJavaScript: true, or else the launch method will not work properly. On iOS, the default behavior is to open all web URLs within the app. Everything else is redirected to the app handler.

Upvotes: 19

Imran Sefat
Imran Sefat

Reputation: 773

using the url_launcher package to do the following:

dependencies:
  url_launcher: ^latest_version


if (await canLaunchUrl(Uri.parse(url))) {
    await launchUrl(Uri.parse(url));
}

Note: Ensure you are trying to open a URI, not a String.

Upvotes: 0

user9723749
user9723749

Reputation:

The PLUGIN plugin works great, as you explain in your examples.

import 'package:flutter/material.dart';
import 'package:url_launcher/url_launcher.dart';
final Uri _url = Uri.parse('https://flutter.dev');
void main() => runApp(
  const MaterialApp(
    home: Material(
      child: Center(
        child: ElevatedButton(
          onPressed: launchUrlStart(url: "https://flutter.dev"),
          child: Text('Show Flutter homepage'),
        ),
      ),
    ),
  ),
);

Future<void> launchUrlStart({required String url}) async {
if (!await launchUrl(Uri.parse(url))) {
  throw 'Could not launch $url';
 }
}

But when trying to open PDF https://www.orimi.com/pdf-test.pdf it remained blank, the problem was that the browser handled it in its own way. Therefore the solution was to tell it to open with an external application and it worked as expected.

Future<void> launchUrlStart({required String url}) async {
  if (!await launchUrl(Uri.parse(url),mode: LaunchMode.externalApplication)) {
    throw 'Could not launch $url';
  }
}

In pubspec.yaml

#https://pub.dev/packages/url_launcher
url_launcher: ^6.1.5

Upvotes: 2

sajid
sajid

Reputation: 215

This is now implemented as Plugin

    const url = "https://flutter.io";
    final Uri _url = Uri.parse(url);

    await launchUrl(_url,mode: LaunchMode.externalApplication);

 pubspec.yaml 
Add dependencies

dependencies: url_launcher: ^6.0.12

Output

Upvotes: 9

Andrew
Andrew

Reputation: 8090

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.

Upvotes: 81

Pai-Hsiang Huang
Pai-Hsiang Huang

Reputation: 372

For those who wants to implement LAUNCH BROWSER AND EXIT APP by using url_launcher. Remember to use (forceSafariVC: false) to open the url in default browser of the phone. Otherwise, the launched browser exit along with your APP.

await launch(URL, forceSafariVC: false);

Upvotes: 14

Rohan Taneja
Rohan Taneja

Reputation: 10667

For Flutter:

As described above by Günter Zöchbauer

For Flutter Web:

import 'dart:html' as html;

Then use:

html.window.open(url, name);

Make sure that you run flutter clean if the import doesn't resolve.

Upvotes: 17

edgar wahl
edgar wahl

Reputation: 51

If you want to use url_launcher than please use it in this form 

environment:
  sdk: ">=2.1.0 <3.0.0"

dependencies:
  url_launcher: ^5.0.2
  flutter:
    sdk: flutter

This answer is also for absolute beginners: They are thinking behind the flutter sdk. No that was a failure. The packages were extras and not in the flutter Sdk. These were secondary packages (single small framework helpers).

Upvotes: 5

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