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);

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

下面的代码可能会帮助你..不要忘记添加权限..

public String getLocalIpAddress(){
   try {
       for (Enumeration<NetworkInterface> en = NetworkInterface.getNetworkInterfaces();  
       en.hasMoreElements();) {
       NetworkInterface intf = en.nextElement();
           for (Enumeration<InetAddress> enumIpAddr = intf.getInetAddresses(); enumIpAddr.hasMoreElements();) {
           InetAddress inetAddress = enumIpAddr.nextElement();
                if (!inetAddress.isLoopbackAddress()) {
                return inetAddress.getHostAddress();
                }
           }
       }
       } catch (Exception ex) {
          Log.e("IP Address", ex.toString());
      }
      return null;
}

在清单文件中添加以下权限。

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

编码快乐! !

private InetAddress getLocalAddress()throws IOException {

            try {
                for (Enumeration<NetworkInterface> en = NetworkInterface.getNetworkInterfaces(); en.hasMoreElements();) {
                    NetworkInterface intf = en.nextElement();
                    for (Enumeration<InetAddress> enumIpAddr = intf.getInetAddresses(); enumIpAddr.hasMoreElements();) {
                        InetAddress inetAddress = enumIpAddr.nextElement();
                        if (!inetAddress.isLoopbackAddress()) {
                            //return inetAddress.getHostAddress().toString();
                            return inetAddress;
                        }
                    }
                }
            } catch (SocketException ex) {
                Log.e("SALMAN", ex.toString());
            }
            return null;
        }

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);

在你的活动中，下面的函数getIpAddress(context)返回电话的IP地址:

public static String getIpAddress(Context context) {
    WifiManager wifiManager = (WifiManager) context.getApplicationContext()
                .getSystemService(WIFI_SERVICE);

    String ipAddress = intToInetAddress(wifiManager.getDhcpInfo().ipAddress).toString();

    ipAddress = ipAddress.substring(1);

    return ipAddress;
}

public static InetAddress intToInetAddress(int hostAddress) {
    byte[] addressBytes = { (byte)(0xff & hostAddress),
                (byte)(0xff & (hostAddress >> 8)),
                (byte)(0xff & (hostAddress >> 16)),
                (byte)(0xff & (hostAddress >> 24)) };

    try {
        return InetAddress.getByAddress(addressBytes);
    } catch (UnknownHostException e) {
        throw new AssertionError();
    }
}