return Dns.GetHostEntry(Dns.GetHostName()).AddressList.FirstOrDefault(ip => ip.AddressFamily == AddressFamily.InterNetwork);

简单的一行代码，返回第一个内部IPV4地址，如果没有则返回null。作为上面的注释添加，但可能对某人有用(上面的一些解决方案将返回多个需要进一步过滤的地址)。

它也很容易返回环回而不是null，我猜用:

return Dns.GetHostEntry(Dns.GetHostName()).AddressList.FirstOrDefault(ip => ip.AddressFamily == AddressFamily.InterNetwork) ?? new IPAddress( new byte[] {127, 0, 0, 1} );

试试这个:

 IPAddress[] localIPs = Dns.GetHostAddresses(Dns.GetHostName());
 String MyIp = localIPs[0].ToString();

以最快的方式获取远程ip地址。您必须使用下载程序，或者在您的计算机上创建一个服务器。

使用这段简单代码的缺点:(这是推荐的)是它将花费3-5秒来获取您的远程IP地址，因为web客户端初始化时总是需要3-5秒来检查您的代理设置。

 public static string GetIP()
 {
            string externalIP = "";
            externalIP = new WebClient().DownloadString("http://checkip.dyndns.org/");
            externalIP = (new Regex(@"\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}"))
                                           .Matches(externalIP)[0].ToString();
            return externalIP;
 }

这是我如何修复它..(第一次仍然需要3-5秒)，但在那之后，它总是会在0-2秒内得到你的远程IP地址，这取决于你的连接。

public static WebClient webclient = new WebClient();
public static string GetIP()
{
    string externalIP = "";
    externalIP = webclient.DownloadString("http://checkip.dyndns.org/");
    externalIP = (new Regex(@"\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}"))
                                   .Matches(externalIP)[0].ToString();
    return externalIP;
}

namespace NKUtilities 
{
    using System;
    using System.Net;

    public class DNSUtility
    {
        public static int Main (string [] args)
        {

          String strHostName = new String ("");
          if (args.Length == 0)
          {
              // Getting Ip address of local machine...
              // First get the host name of local machine.
              strHostName = Dns.GetHostName ();
              Console.WriteLine ("Local Machine's Host Name: " +  strHostName);
          }
          else
          {
              strHostName = args[0];
          }

          // Then using host name, get the IP address list..
          IPHostEntry ipEntry = DNS.GetHostByName (strHostName);
          IPAddress [] addr = ipEntry.AddressList;

          for (int i = 0; i < addr.Length; i++)
          {
              Console.WriteLine ("IP Address {0}: {1} ", i, addr[i].ToString ());
          }
          return 0;
        }    
     }
}

为了找到IP地址列表，我已经使用了这个解决方案

public static IEnumerable<string> GetAddresses()
{
    var host = Dns.GetHostEntry(Dns.GetHostName());
    return (from ip in host.AddressList where ip.AddressFamily == AddressFamily.lo select ip.ToString()).ToList();
}

但我个人喜欢下面的解决方案，以获得本地有效的IP地址

public static IPAddress GetIPAddress(string hostName)
{
    Ping ping = new Ping();
    var replay = ping.Send(hostName);

    if (replay.Status == IPStatus.Success)
    {
        return replay.Address;
    }
    return null;
 }

public static void Main()
{
    Console.WriteLine("Local IP Address: " + GetIPAddress(Dns.GetHostName()));
    Console.WriteLine("Google IP:" + GetIPAddress("google.com");
    Console.ReadLine();
}

清洁和一体解决方案:D

//This returns the first IP4 address or null
return Dns.GetHostEntry(Dns.GetHostName()).AddressList.FirstOrDefault(ip => ip.AddressFamily == AddressFamily.InterNetwork);