虽然有一个正确的答案，我在这里分享我的答案，希望这样会更方便。

WifiManager wifiMan = (WifiManager) context.getSystemService(Context.WIFI_SERVICE);
WifiInfo wifiInf = wifiMan.getConnectionInfo();
int ipAddress = wifiInf.getIpAddress();
String ip = String.format("%d.%d.%d.%d", (ipAddress & 0xff),(ipAddress >> 8 & 0xff),(ipAddress >> 16 & 0xff),(ipAddress >> 24 & 0xff));

老实说，我对代码安全只是有点熟悉，所以这可能有点像黑客。但对我来说，这是最通用的方法:

package com.my_objects.ip;

import java.net.InetAddress;
import java.net.UnknownHostException;

public class MyIpByHost 
{
  public static void main(String a[])
  {
   try 
    {
      InetAddress host = InetAddress.getByName("nameOfDevice or webAddress");
      System.out.println(host.getHostAddress());
    } 
   catch (UnknownHostException e) 
    {
      e.printStackTrace();
    }
} }

如果你有一个壳;Ifconfig eth0也适用于x86设备

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。

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) {}
   }
}

我不使用Android，但我会用完全不同的方式来解决这个问题。

发送一个查询到谷歌，像这样: https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=my%20ip

并引用发布响应的HTML字段。您也可以直接查询到源。

谷歌最可能比你的应用程序存在的时间长。

只要记住，这可能是你的用户在这个时候没有互联网，你希望发生什么!

祝你好运