在互联网上有几个地方告诉你如何获得一个IP地址。很多都像这个例子:

String strHostName = string.Empty;
// 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);
// Then using host name, get the IP address list..
IPHostEntry ipEntry = Dns.GetHostEntry(strHostName);
IPAddress[] addr = ipEntry.AddressList;

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

在这个例子中,我得到了几个IP地址,但我只对路由器分配给运行程序的计算机的IP地址感兴趣:例如,如果某人希望访问我计算机中的共享文件夹,我将给他的IP地址。

如果我没有连接到网络,我直接通过没有路由器的调制解调器连接到互联网,那么我希望得到一个错误。我如何才能看到我的计算机是否连接到网络与c#,如果它是然后得到局域网IP地址。


当前回答

这将从两个单独的列表(IPV4和IPV6)中具有网关地址和单播地址的任何接口返回地址。

public static (List<IPAddress> V4, List<IPAddress> V6) GetLocal()
{
    List<IPAddress> foundV4 = new List<IPAddress>();
    List<IPAddress> foundV6 = new List<IPAddress>();

    NetworkInterface.GetAllNetworkInterfaces().ToList().ForEach(ni =>
    {
        if (ni.GetIPProperties().GatewayAddresses.FirstOrDefault() != null)
        {
            ni.GetIPProperties().UnicastAddresses.ToList().ForEach(ua =>
            {
                if (ua.Address.AddressFamily == AddressFamily.InterNetwork) foundV4.Add(ua.Address);
                if (ua.Address.AddressFamily == AddressFamily.InterNetworkV6) foundV6.Add(ua.Address);
            });
        }
    });

    return (foundV4.Distinct().ToList(), foundV6.Distinct().ToList());
}

其他回答

我认为使用LINQ更容易:

Dns.GetHostEntry(Dns.GetHostName())
   .AddressList
   .First(x => x.AddressFamily == System.Net.Sockets.AddressFamily.InterNetwork)
   .ToString()

我还在努力获取正确的IP。

我在这里尝试了各种解决方案,但没有一个能达到我想要的效果。几乎所有提供的条件测试都没有使用地址。

这是对我有效的方法,希望能有所帮助……

var firstAddress = (from address in NetworkInterface.GetAllNetworkInterfaces().Select(x => x.GetIPProperties()).SelectMany(x => x.UnicastAddresses).Select(x => x.Address)
                    where !IPAddress.IsLoopback(address) && address.AddressFamily == System.Net.Sockets.AddressFamily.InterNetwork
                    select address).FirstOrDefault();

Console.WriteLine(firstAddress);

前提条件:你必须添加System.Data.Linq引用并引用它

using System.Linq;
string ipAddress ="";
IPHostEntry ipHostInfo = Dns.GetHostEntry(Dns.GetHostName());
ipAddress = Convert.ToString(ipHostInfo.AddressList.FirstOrDefault(address => address.AddressFamily == AddressFamily.InterNetwork));

这是最短的方法:

Dns.GetHostEntry(
    Dns.GetHostName()
).AddressList.AsEnumerable().Where(
    ip=>ip.AddressFamily.Equals(AddressFamily.InterNetwork)
).FirstOrDefault().ToString()

使用一个或多个LAN卡和虚拟机进行测试

public static string DisplayIPAddresses()
    {
        string returnAddress = String.Empty;

        // Get a list of all network interfaces (usually one per network card, dialup, and VPN connection)
        NetworkInterface[] networkInterfaces = NetworkInterface.GetAllNetworkInterfaces();

        foreach (NetworkInterface network in networkInterfaces)
        {
            // Read the IP configuration for each network
            IPInterfaceProperties properties = network.GetIPProperties();

            if (network.NetworkInterfaceType == NetworkInterfaceType.Ethernet &&
                   network.OperationalStatus == OperationalStatus.Up &&
                   !network.Description.ToLower().Contains("virtual") &&
                   !network.Description.ToLower().Contains("pseudo"))
            {
                // Each network interface may have multiple IP addresses
                foreach (IPAddressInformation address in properties.UnicastAddresses)
                {
                    // We're only interested in IPv4 addresses for now
                    if (address.Address.AddressFamily != AddressFamily.InterNetwork)
                        continue;

                    // Ignore loopback addresses (e.g., 127.0.0.1)
                    if (IPAddress.IsLoopback(address.Address))
                        continue;

                    returnAddress = address.Address.ToString();
                    Console.WriteLine(address.Address.ToString() + " (" + network.Name + " - " + network.Description + ")");
                }
            }
        }

       return returnAddress;
    }