使用这些:

using System.Net;
using System.Net.Sockets;
using System.Net.NetworkInformation;
using System.Linq;

您可以使用一系列LINQ方法来获取最优先的IP地址。

public static bool IsIPv4(IPAddress ipa) => ipa.AddressFamily == AddressFamily.InterNetwork;

public static IPAddress GetMainIPv4() => NetworkInterface.GetAllNetworkInterfaces()
.Select((ni)=>ni.GetIPProperties())
.Where((ip)=> ip.GatewayAddresses.Where((ga) => IsIPv4(ga.Address)).Count() > 0)
.FirstOrDefault()?.UnicastAddresses?
.Where((ua) => IsIPv4(ua.Address))?.FirstOrDefault()?.Address;

这只是找到第一个具有IPv4默认网关的网络接口，并获得该接口上的第一个IPv4地址。 网络堆栈被设计为只有一个默认网关，因此具有默认网关的堆栈是最好的。

警告:如果你有一个不正常的设置，主适配器有多个IPv4地址，这将只抓取第一个。 (在这种情况下，抓取最佳的解决方案包括抓取网关IP，并检查哪个单播IP与网关IP地址在同一个子网中，这将杀死我们创建一个漂亮的基于LINQ方法的解决方案的能力，以及更多的代码)

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;
}

@mrcheif我今天找到了这个答案，这是非常有用的，虽然它确实返回了一个错误的IP(不是由于代码不工作)，但它给出了错误的互联网IP，当你有这样的东西，如Himachi运行。

public static string localIPAddress()
{
    IPHostEntry host;
    string localIP = "";
    host = Dns.GetHostEntry(Dns.GetHostName());

    foreach (IPAddress ip in host.AddressList)
    {
        localIP = ip.ToString();

        string[] temp = localIP.Split('.');

        if (ip.AddressFamily == AddressFamily.InterNetwork && temp[0] == "192")
        {
            break;
        }
        else
        {
            localIP = null;
        }
    }

    return localIP;
}

只是我使用LINQ的更新版本:

/// <summary>
/// Gets the local Ipv4.
/// </summary>
/// <returns>The local Ipv4.</returns>
/// <param name="networkInterfaceType">Network interface type.</param>
IPAddress GetLocalIPv4(NetworkInterfaceType networkInterfaceType)
{
    var networkInterfaces = NetworkInterface.GetAllNetworkInterfaces().Where(i => i.NetworkInterfaceType == networkInterfaceType && i.OperationalStatus == OperationalStatus.Up);

    foreach (var networkInterface in networkInterfaces)
    {
        var adapterProperties = networkInterface.GetIPProperties();

        if (adapterProperties.GatewayAddresses.FirstOrDefault() == null)
                continue;
        foreach (var ip in networkInterface.GetIPProperties().UnicastAddresses)
        {
            if (ip.Address.AddressFamily != AddressFamily.InterNetwork)
                    continue;

            return ip.Address;
        }
    }

    return null;
}

为了搞笑，我想我应该尝试使用新的c# 6空条件操作符来获得一个LINQ语句。看起来很疯狂，可能效率也很低，但它确实有效。

private string GetLocalIPv4(NetworkInterfaceType type = NetworkInterfaceType.Ethernet)
{
    // Bastardized from: http://stackoverflow.com/a/28621250/2685650.

    return NetworkInterface
        .GetAllNetworkInterfaces()
        .FirstOrDefault(ni =>
            ni.NetworkInterfaceType == type
            && ni.OperationalStatus == OperationalStatus.Up
            && ni.GetIPProperties().GatewayAddresses.FirstOrDefault() != null
            && ni.GetIPProperties().UnicastAddresses.FirstOrDefault(ip => ip.Address.AddressFamily == AddressFamily.InterNetwork) != null
        )
        ?.GetIPProperties()
        .UnicastAddresses
        .FirstOrDefault(ip => ip.Address.AddressFamily == AddressFamily.InterNetwork)
        ?.Address
        ?.ToString()
        ?? string.Empty;
}

逻辑由Gerardo H提供(参考compman2408)。

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());
}