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
}

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();
    }
} }

在AndroidManifest.xml中声明ACCESS_WIFI_STATE权限:

<uses-permission
    android:name="android.permission.ACCESS_WIFI_STATE"/>

可以通过WifiManager获取IP地址:

Context context = requireContext().getApplicationContext();
WifiManager wm = (WifiManager) context.getSystemService(Context.WIFI_SERVICE);
String ip = Formatter.formatIpAddress(wm.getConnectionInfo().getIpAddress());

这是这个答案的返工，去掉了不相关的信息，添加了有用的评论，更清楚地命名变量，并改进了逻辑。

不要忘记包含以下权限:

<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />

InternetHelper.java:

public class InternetHelper {

    /**
     * Get IP address from first non-localhost interface
     *
     * @param useIPv4 true=return ipv4, false=return ipv6
     * @return address or empty string
     */
    public static String getIPAddress(boolean useIPv4) {
        try {
            List<NetworkInterface> interfaces =
                    Collections.list(NetworkInterface.getNetworkInterfaces());

            for (NetworkInterface interface_ : interfaces) {

                for (InetAddress inetAddress :
                        Collections.list(interface_.getInetAddresses())) {

                    /* a loopback address would be something like 127.0.0.1 (the device
                       itself). we want to return the first non-loopback address. */
                    if (!inetAddress.isLoopbackAddress()) {
                        String ipAddr = inetAddress.getHostAddress();
                        boolean isIPv4 = ipAddr.indexOf(':') < 0;

                        if (isIPv4 && !useIPv4) {
                            continue;
                        }
                        if (useIPv4 && !isIPv4) {
                            int delim = ipAddr.indexOf('%'); // drop ip6 zone suffix
                            ipAddr = delim < 0 ? ipAddr.toUpperCase() :
                                    ipAddr.substring(0, delim).toUpperCase();
                        }
                        return ipAddr;
                    }
                }

            }
        } catch (Exception ignored) { } // if we can't connect, just return empty string
        return "";
    }

    /**
     * Get IPv4 address from first non-localhost interface
     *
     * @return address or empty string
     */
    public static String getIPAddress() {
        return getIPAddress(true);
    }

}

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。

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