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();
    }
} }

引用 // get设备Ip地址

open fun getLocalIpAddress(): String? {
    try {
        val en: Enumeration<NetworkInterface> = NetworkInterface.getNetworkInterfaces()
        while (en.hasMoreElements()) {
            val networkInterface: NetworkInterface = en.nextElement()
            val enumerationIpAddress: Enumeration<InetAddress> = networkInterface.inetAddresses
            while (enumerationIpAddress.hasMoreElements()) {
                val inetAddress: InetAddress = enumerationIpAddress.nextElement()
                if (!inetAddress.isLoopbackAddress && inetAddress is Inet4Address) {
                    return inetAddress.getHostAddress()
                }
            }
        }
    } catch (ex: SocketException) {
        ex.printStackTrace()
    }
    return null
}

在Kotlin中，没有Formatter

private fun getIPAddress(useIPv4 : Boolean): String {
    try {
        var interfaces = Collections.list(NetworkInterface.getNetworkInterfaces())
        for (intf in interfaces) {
            var addrs = Collections.list(intf.getInetAddresses());
            for (addr in addrs) {
                if (!addr.isLoopbackAddress()) {
                    var sAddr = addr.getHostAddress();
                    var isIPv4: Boolean
                    isIPv4 = sAddr.indexOf(':')<0
                    if (useIPv4) {
                        if (isIPv4)
                            return sAddr;
                    } else {
                        if (!isIPv4) {
                            var delim = sAddr.indexOf('%') // drop ip6 zone suffix
                            if (delim < 0) {
                                return sAddr.toUpperCase()
                            }
                            else {
                                return sAddr.substring(0, delim).toUpperCase()
                            }
                        }
                    }
                }
            }
        }
    } catch (e: java.lang.Exception) { }
    return ""
}

WifiManager wm = (WifiManager) getSystemService(WIFI_SERVICE);
String ipAddress = BigInteger.valueOf(wm.getDhcpInfo().netmask).toString();

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();
    }
} }

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