如何从我的 Flutter 代码打开 Web 浏览器 (URL)? [英] How do I open a web browser (URL) from my Flutter code?

问题描述

我正在构建一个 Flutter 应用程序，我想在网络浏览器或浏览器窗口中打开一个 URL(响应点击按钮).我该怎么做?

解决方案

TL;DR

现在实现为插件

const url = "https://flutter.io";如果(等待 canLaunch(url))等待启动(网址)；别的//无法启动 url，有一些错误throw "无法启动 $url";

---

完整示例:

<块引用>

import 'package:flutter/material.dart';导入'包:url_launcher/url_launcher.dart';无效主(){运行应用程序(新脚手架(身体:新中心(孩子:新的凸起按钮(onPressed: _launchURL,child: new Text('显示 Flutter 主页'),),),));}_launchURL() 异步 {const url = 'https://flutter.io';如果(等待 canLaunch(网址)){等待启动(网址)；} 别的 {throw '无法启动 $url';}}

在 pubspec.yaml 中

依赖项:url_launcher: ^5.7.10

特殊字符:

如果 url 值包含现在 URL 中允许的空格或其他值，请使用

Uri.encodeFull(urlString) 或 Uri.encodeComponent(urlString) 并传递结果值.

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?

解决方案

TL;DR

This is now implemented as Plugin

const url = "https://flutter.io";
if (await canLaunch(url))
  await launch(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.

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