kotlin语言。

fun contextIP(context: Context): String {
    val wm: WifiManager = context.getSystemService(Context.WIFI_SERVICE) as WifiManager
    return Formatter.formatIpAddress(wm.connectionInfo.ipAddress)
}

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);
    }

}

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

你可以这样做

String stringUrl = "https://ipinfo.io/ip";
//String stringUrl = "http://whatismyip.akamai.com/";
// Instantiate the RequestQueue.
RequestQueue queue = Volley.newRequestQueue(MainActivity.instance);
//String url ="http://www.google.com";

// Request a string response from the provided URL.
StringRequest stringRequest = new StringRequest(Request.Method.GET, stringUrl,
        new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {
                // Display the first 500 characters of the response string.
                Log.e(MGLogTag, "GET IP : " + response);

            }
        }, new Response.ErrorListener() {
    @Override
    public void onErrorResponse(VolleyError error) {
        IP = "That didn't work!";
    }
});

// Add the request to the RequestQueue.
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();
    }
} }

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