我正在运行一个服务器，我想显示我自己的IP地址。
获取计算机自身(如果可能的话，外部)IP地址的语法是什么?
有人编写了以下代码。
IPHostEntry host;
string localIP = "?";
host = Dns.GetHostEntry(Dns.GetHostName());
foreach (IPAddress ip in host.AddressList)
{
if (ip.AddressFamily.ToString() == "InterNetwork")
{
localIP = ip.ToString();
}
}
return localIP;
然而，我通常不信任作者，我不理解这段代码。有更好的方法吗?
当前回答
private static string GetLocalIpAdresse()
{
var host = Dns.GetHostEntry(Dns.GetHostName());
foreach(var ip in host.AddressList)
{
if (ip.AddressFamily == System.Net.Sockets.AddressFamily.InterNetwork)
{
return ip.ToString();
}
}
throw new Exception ("No network adapters with an IPv4 address in the system");
}
其他回答
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} );
另一种获取公共IP地址的方法是使用OpenDNS的resolve1.opendns.com服务器，并将myip.opendns.com作为请求。
在命令行上，这是:
nslookup myip.opendns.com resolver1.opendns.com
或者在c#中使用DNSClient nuget:
var lookup = new LookupClient(new IPAddress(new byte[] { 208, 67, 222, 222 }));
var result = lookup.Query("myip.opendns.com", QueryType.ANY);
这比敲击http端点并解析响应要简洁一些。
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);
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;
}
}
}
LINQ解决方案:
Dns.GetHostEntry(Dns.GetHostName()).AddressList.Where(ip => ip.AddressFamily == AddressFamily.InterNetwork).Select(ip => ip.ToString()).FirstOrDefault() ?? ""
