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

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);

其他回答

用Linq更新Mrchief的答案,我们会有:

public static IPAddress GetLocalIPAddress()
{
   var host = Dns.GetHostEntry(Dns.GetHostName());
   var ipAddress= host.AddressList.FirstOrDefault(ip => ip.AddressFamily == AddressFamily.InterNetwork);
   return ipAddress;
}

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;
    }

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));

已经有很多答案了,但我还是贡献我的一个:

public static IPAddress LocalIpAddress() {
    Func<IPAddress, bool> localIpPredicate = ip =>
        ip.AddressFamily == AddressFamily.InterNetwork &&
        ip.ToString().StartsWith("192.168"); //check only for 16-bit block
    return Dns.GetHostEntry(Dns.GetHostName()).AddressList.LastOrDefault(localIpPredicate);
}

一个衬套:

public static IPAddress LocalIpAddress() => Dns.GetHostEntry(Dns.GetHostName()).AddressList.LastOrDefault(ip => ip.AddressFamily == AddressFamily.InterNetwork && ip.ToString().StartsWith("192.168"));

注意:从last开始搜索,因为它在一些接口添加到设备后仍然有效,如MobileHotspot,VPN或其他花哨的虚拟适配器。

过时了,这对我有用

public static IPAddress GetIPAddress()
{ 
 IPAddress ip = Dns.GetHostAddresses(Dns.GetHostName()).Where(address => 
 address.AddressFamily == AddressFamily.InterNetwork).First();
 return ip;
}