Kotlin极简版

fun getIpv4HostAddress(): String {
    NetworkInterface.getNetworkInterfaces()?.toList()?.map { networkInterface ->
        networkInterface.inetAddresses?.toList()?.find {
            !it.isLoopbackAddress && it is Inet4Address
        }?.let { return it.hostAddress }
    }
    return ""
}

如果你有一个壳;Ifconfig eth0也适用于x86设备

根据我的测试，这是我的建议

import java.net.*;
import java.util.*;

public class hostUtil
{
   public static String HOST_NAME = null;
   public static String HOST_IPADDRESS = null;

   public static String getThisHostName ()
   {
      if (HOST_NAME == null) obtainHostInfo ();
      return HOST_NAME;
   }

   public static String getThisIpAddress ()
   {
      if (HOST_IPADDRESS == null) obtainHostInfo ();
      return HOST_IPADDRESS;
   }

   protected static void obtainHostInfo ()
   {
      HOST_IPADDRESS = "127.0.0.1";
      HOST_NAME = "localhost";

      try
      {
         InetAddress primera = InetAddress.getLocalHost();
         String hostname = InetAddress.getLocalHost().getHostName ();

         if (!primera.isLoopbackAddress () &&
             !hostname.equalsIgnoreCase ("localhost") &&
              primera.getHostAddress ().indexOf (':') == -1)
         {
            // Got it without delay!!
            HOST_IPADDRESS = primera.getHostAddress ();
            HOST_NAME = hostname;
            //System.out.println ("First try! " + HOST_NAME + " IP " + HOST_IPADDRESS);
            return;
         }
         for (Enumeration<NetworkInterface> netArr = NetworkInterface.getNetworkInterfaces(); netArr.hasMoreElements();)
         {
            NetworkInterface netInte = netArr.nextElement ();
            for (Enumeration<InetAddress> addArr = netInte.getInetAddresses (); addArr.hasMoreElements ();)
            {
               InetAddress laAdd = addArr.nextElement ();
               String ipstring = laAdd.getHostAddress ();
               String hostName = laAdd.getHostName ();

               if (laAdd.isLoopbackAddress()) continue;
               if (hostName.equalsIgnoreCase ("localhost")) continue;
               if (ipstring.indexOf (':') >= 0) continue;

               HOST_IPADDRESS = ipstring;
               HOST_NAME = hostName;
               break;
            }
         }
      } catch (Exception ex) {}
   }
}

一个设备可能有几个IP地址，在一个特定的应用程序中使用的IP地址可能不是接收请求的服务器将看到的IP地址。事实上，一些用户使用VPN或Cloudflare Warp等代理。

如果你的目的是获得IP地址，就像服务器从你的设备接收请求一样，那么最好是通过Java客户端查询IP地理定位服务，如Ipregistry(免责声明:我为该公司工作):

https://github.com/ipregistry/ipregistry-java

IpregistryClient client = new IpregistryClient("tryout");
RequesterIpInfo requesterIpInfo = client.lookup();
requesterIpInfo.getIp();

除了使用非常简单之外，您还可以获得其他信息，例如国家、语言、货币、设备IP的时区，并且您可以识别用户是否正在使用代理。

Kotlin极简版

fun getIpv4HostAddress(): String {
    NetworkInterface.getNetworkInterfaces()?.toList()?.map { networkInterface ->
        networkInterface.inetAddresses?.toList()?.find {
            !it.isLoopbackAddress && it is Inet4Address
        }?.let { return it.hostAddress }
    }
    return ""
}

您不需要像目前提供的解决方案那样添加权限。以字符串形式下载此网站:

http://www.ip-api.com/json

or

http://www.telize.com/geoip

下载一个网站作为字符串可以用java代码完成:

http://www.itcuties.com/java/read-url-to-string/

像这样解析JSON对象:

https://stackoverflow.com/a/18998203/1987258

json属性“query”或“ip”包含ip地址。