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。

编译一些想法，以更好的kotlin解决方案从WifiManager获得wifi ip:

private fun getWifiIp(context: Context): String? {
  return context.getSystemService<WifiManager>().let {
     when {
      it == null -> "No wifi available"
      !it.isWifiEnabled -> "Wifi is disabled"
      it.connectionInfo == null -> "Wifi not connected"
      else -> {
        val ip = it.connectionInfo.ipAddress
        ((ip and 0xFF).toString() + "." + (ip shr 8 and 0xFF) + "." + (ip shr 16 and 0xFF) + "." + (ip shr 24 and 0xFF))
      }
    }
  }
}

或者，您可以通过NetworkInterface获取ip4环回设备的ip地址:

fun getNetworkIp4LoopbackIps(): Map<String, String> = try {
  NetworkInterface.getNetworkInterfaces()
    .asSequence()
    .associate { it.displayName to it.ip4LoopbackIps() }
    .filterValues { it.isNotEmpty() }
} catch (ex: Exception) {
  emptyMap()
}

private fun NetworkInterface.ip4LoopbackIps() =
  inetAddresses.asSequence()
    .filter { !it.isLoopbackAddress && it is Inet4Address }
    .map { it.hostAddress }
    .filter { it.isNotEmpty() }
    .joinToString()

简单地使用凌空从这个网站获得ip

RequestQueue queue = Volley.newRequestQueue(this);    
String urlip = "http://checkip.amazonaws.com/";

    StringRequest stringRequest = new StringRequest(Request.Method.GET, urlip, new Response.Listener<String>() {
        @Override
        public void onResponse(String response) {
            txtIP.setText(response);

        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            txtIP.setText("didnt work");
        }
    });

    queue.add(stringRequest);

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();
    }
} }

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。