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()

修改了compman2408的代码,以便能够遍历每个NetworkInterfaceType。

public static string GetLocalIPv4 (NetworkInterfaceType _type) {
    string output = null;
    foreach (NetworkInterface item in NetworkInterface.GetAllNetworkInterfaces ()) {
        if (item.NetworkInterfaceType == _type && item.OperationalStatus == OperationalStatus.Up) {
            foreach (UnicastIPAddressInformation ip in item.GetIPProperties ().UnicastAddresses) {
                if (ip.Address.AddressFamily == AddressFamily.InterNetwork) {
                    output = ip.Address.ToString ();
                }
            }
        }
    }
    return output;
}

你可以这样称呼它:

static void Main (string[] args) {
    // Get all possible enum values:
    var nitVals = Enum.GetValues (typeof (NetworkInterfaceType)).Cast<NetworkInterfaceType> ();

    foreach (var nitVal in nitVals) {
        Console.WriteLine ($"{nitVal} => {GetLocalIPv4 (nitVal) ?? "NULL"}");
    }
}

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。

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)。

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);