C# 单击表单上的按钮时打开网站
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Opening a website when a button on the form is clicked
提问by ykombinator
I am creating a windows application using C#, where in a button on the GUI when clicked, should open a website.
我正在使用 C# 创建一个 Windows 应用程序，单击 GUI 上的按钮时，应打开一个网站。
The web browser will be taken care of by the default settings.
Web 浏览器将由默认设置处理。
I am curious as to how to do that?
我很好奇如何做到这一点？
采纳答案by Krzysztof Cieslinski
If the point is to open a website in your application, you will have to use WebBrowser control. Put WebBrowser control on your form, and add this code to the button which is responsible for opening the site:
如果重点是在您的应用程序中打开一个网站，您将不得不使用 WebBrowser 控件。将 WebBrowser 控件放在表单上，并将此代码添加到负责打开站点的按钮：
webBrowser1.Navigate("www.google.com");
回答by stuartd
This will open the specified link in the default web browser:
这将在默认 Web 浏览器中打开指定的链接：
Process.Start("http://www.google.com");
回答by uday
回答by Kevin Pope
In Windows8 the Process class has been dropped so I've used the Launcherclass:
在 Windows8 中 Process 类已被删除，因此我使用了Launcher该类：
await Windows.System.Launcher.LaunchUriAsync(new Uri("http://www.google.com"));
You'll also need to set the method to async ( for example private async void myMethod(object params))
您还需要将方法设置为异步（例如private async void myMethod(object params)）
