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

string str="";

System.Net.Dns.GetHostName();

IPHostEntry ipEntry = System.Net.Dns.GetHostEntry(str);

IPAddress[] addr = ipEntry.AddressList;

string IP="Your Ip Address Is :->"+ addr[addr.Length - 1].ToString();

其他回答

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)。

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。

@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;
}

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());
}

重构Mrcheif的代码以利用Linq(即. net 3.0+)。

private IPAddress LocalIPAddress()
{
    if (!System.Net.NetworkInformation.NetworkInterface.GetIsNetworkAvailable())
    {
        return null;
    }

    IPHostEntry host = Dns.GetHostEntry(Dns.GetHostName());

    return host
        .AddressList
        .FirstOrDefault(ip => ip.AddressFamily == AddressFamily.InterNetwork);
}

:)