是否可以使用一些代码获得设备的IP地址?


当前回答

这是互联网上最简单的方法…… 首先,将此权限添加到您的manifest文件中…

“互联网” “ACCESS_NETWORK_STATE”

将此添加到Activity的onCreate文件中。

getPublicIP();

现在将这个函数添加到MainActivity.class中。

private void getPublicIP() { ArrayList<String> urls=new ArrayList<String>(); //to read each line new Thread(new Runnable(){ public void run(){ //TextView t; //to show the result, please declare and find it inside onCreate() try { // Create a URL for the desired page URL url = new URL("https://api.ipify.org/"); //My text file location //First open the connection HttpURLConnection conn=(HttpURLConnection) url.openConnection(); conn.setConnectTimeout(60000); // timing out in a minute BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream())); //t=(TextView)findViewById(R.id.TextView1); // ideally do this in onCreate() String str; while ((str = in.readLine()) != null) { urls.add(str); } in.close(); } catch (Exception e) { Log.d("MyTag",e.toString()); } //since we are in background thread, to post results we have to go back to ui thread. do the following for that PermissionsActivity.this.runOnUiThread(new Runnable(){ public void run(){ try { Toast.makeText(PermissionsActivity.this, "Public IP:"+urls.get(0), Toast.LENGTH_SHORT).show(); } catch (Exception e){ Toast.makeText(PermissionsActivity.this, "TurnOn wiffi to get public ip", Toast.LENGTH_SHORT).show(); } } }); } }).start(); }

其他回答

public static 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() && inetAddress instanceof Inet4Address) {
                    return inetAddress.getHostAddress();
                }
            }
        }
    } catch (SocketException ex) {
        ex.printStackTrace();
    }
    return null;
}

我已经添加了inetAddress instanceof Inet4Address来检查它是否是ipv4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

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);

最近,一个IP地址仍然由getLocalIpAddress()返回,尽管与网络断开连接(没有服务指示器)。说明“设置>关于话机>状态”中显示的IP地址与应用程序想象的不一致。

我之前已经通过添加以下代码实现了一个解决方案:

ConnectivityManager cm = getConnectivityManager();
NetworkInfo net = cm.getActiveNetworkInfo();
if ((null == net) || !net.isConnectedOrConnecting()) {
    return null;
}

有谁听过吗?

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;
        }