Question

我试过这段代码公共类MainActivity扩展了Activity {

private static final String TAG = null;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    TextView tv= (TextView) findViewById(R.id.textView2);
    String a=getLocalIpAddress();
      tv.setText(a);

}


@Override
public boolean onCreateOptionsMenu(Menu menu) {
    // Inflate the menu; this adds items to the action bar if it is present.
    getMenuInflater().inflate(R.menu.main, menu);
    return true;
}
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()) {
                     String a= inetAddress.getHostAddress();

                     return a;
                    }
                }
            }
        } catch (SocketException ex) {
           // Log.e(tag, ex.toString());
        }
        return "";
    }
}

这给了我像00ff这样的ip地址：22ff：.... 但我希望它像192.168.3.3 ... 格式ip地址已折旧有没有其他方法，以便我可以获得格式化地址

Answer 1

您获得的地址是IPv6。你想要得到的是IPv4，所以尝试使用：

if ((!inetAddress.isLoopbackAddress()) && (InetAddressUtils.isIPv4Address(ipv4 = inetAddress.getHostAddress()))) { ... }

而不是：

if (!inetAddress.isLoopbackAddress()) { ... }

获取自己设备android的格式化IP地址

1 个答案: