根据@ChaosPandion的回答，为了尽可能确保结果是正确的，你可以像其他人指出的那样，包括多个大网站。但是，这应该异步完成，以避免太长的等待时间。此外，WebRequest、HttpWebRequest和HttpWebResponse类现在已经过时，应该被HttpClient取代。以下示例考虑了上述因素:

public static async Task<bool> CheckForInternetConnection(TimeSpan? timeoutMs = null, List<string> urls = null)
{
    if (timeoutMs == null)
    {
        timeoutMs = TimeSpan.FromSeconds(10);
    }

    var culture = CultureInfo.InstalledUICulture;
    if (urls == null)
    {
        urls = new List<string>();

        if (culture.Name.StartsWith("fa"))      // Iran
            urls.Add("http://www.aparat.com");
        else if (culture.Name.StartsWith("zh")) // China
            urls.Add("http://www.baidu.com");
        else
        {
            urls.Add("https://www.apple.com/");
            urls.Add("https://www.gstatic.com/generate_204");
        }
    }

    var client = new HttpClient();
    client.Timeout = (TimeSpan)timeoutMs;
    List<Task<string>> tasks = new List<Task<string>>();
    int unresponsiveUrlCount = 0;

    foreach (var url in urls)
    {   
        tasks.Add(client.GetStringAsync(url));             
    }

    Task aggregationTask = null;
    try
    {
        aggregationTask = Task.WhenAll(tasks);
        await aggregationTask;
    }
    catch (Exception)
    {  
        if (aggregationTask?.Exception?.InnerExceptions != null && aggregationTask.Exception.InnerExceptions.Any())
        {
            foreach (var innerEx in aggregationTask.Exception.InnerExceptions)
            {
                unresponsiveUrlCount++;
            }
        }      
    }

    return unresponsiveUrlCount < urls.Count;
}

该方法检查列表中的所有url，如果它们都不可访问，则返回false。我已经添加了苹果的url，因为在我的情况下，它加载相当快，但它可以替换为任何url。

public static bool CheckForInternetConnection(int timeoutMs = 10000, string url = null)
{
    try
    {
        url ??= CultureInfo.InstalledUICulture switch
        {
            { Name: var n } when n.StartsWith("fa") => // Iran
                "http://www.aparat.com",
            { Name: var n } when n.StartsWith("zh") => // China
                "http://www.baidu.com",
            _ =>
                "http://www.gstatic.com/generate_204",
        };

        var request = (HttpWebRequest)WebRequest.Create(url);
        request.KeepAlive = false;
        request.Timeout = timeoutMs;
        using (var response = (HttpWebResponse)request.GetResponse())
            return true;
    }
    catch
    {
        return false;
    }
}

另一种选择是网络列表管理器API，它适用于Vista和Windows 7。MSDN文章在这里。在本文中有一个下载代码示例的链接，允许您这样做:

AppNetworkListUser nlmUser = new AppNetworkListUser();
Console.WriteLine("Is the machine connected to internet? " + nlmUser.NLM.IsConnectedToInternet.ToString());

请确保从COM选项卡中添加对网络列表1.0类型库的引用…它将显示为NETWORKLIST。

使用NetworkMonitor监控网络状态和internet连接。

示例:

namespace AmRoNetworkMonitor.Demo
{
    using System;

    internal class Program
    {
        private static void Main()
        {
            NetworkMonitor.StateChanged += NetworkMonitor_StateChanged;
            NetworkMonitor.StartMonitor();

            Console.WriteLine("Press any key to stop monitoring.");
            Console.ReadKey();
            NetworkMonitor.StopMonitor();

            Console.WriteLine("Press any key to close program.");
            Console.ReadKey();
        }

        private static void NetworkMonitor_StateChanged(object sender, StateChangeEventArgs e)
        {
            Console.WriteLine(e.IsAvailable ? "Is Available" : "Is Not Available");
        }
    }
}

bool bb = System.Net.NetworkInformation.NetworkInterface.GetIsNetworkAvailable();

if (bb == true)
    MessageBox.Show("Internet connections are available");
else
    MessageBox.Show("Internet connections are not available");

你绝对没有办法可靠地检查是否有互联网连接(我想你的意思是访问互联网)。

但是，您可以请求几乎从未离线的资源，例如ping google.com或类似的东西。我认为这是有效的。

try { 
    Ping myPing = new Ping();
    String host = "google.com";
    byte[] buffer = new byte[32];
    int timeout = 1000;
    PingOptions pingOptions = new PingOptions();
    PingReply reply = myPing.Send(host, timeout, buffer, pingOptions);
    return (reply.Status == IPStatus.Success);
}
catch (Exception) {
    return false;
}