是否可以使用一些代码获得设备的IP地址?


当前回答

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();
    }
}

其他回答

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);

你可以使用LinkProperties。建议用于新的Android版本。

此功能检索WiFi和移动数据的本地IP地址。它需要Manifest.permission。ACCESS_NETWORK_STATE许可。

@Nullable
public static String getDeviceIpAddress(@NonNull ConnectivityManager connectivityManager) {
    LinkProperties linkProperties = connectivityManager.getLinkProperties(connectivityManager.getActiveNetwork());
    InetAddress inetAddress;
    for(LinkAddress linkAddress : linkProperties.getLinkAddresses()) {
        inetAddress = linkAddress.getAddress();
        if (inetAddress instanceof Inet4Address
                && !inetAddress.isLoopbackAddress()
                && inetAddress.isSiteLocalAddress()) {
            return inetAddress.getHostAddress();
        }
    }
    return null;
}

虽然有一个正确的答案,我在这里分享我的答案,希望这样会更方便。

WifiManager wifiMan = (WifiManager) context.getSystemService(Context.WIFI_SERVICE);
WifiInfo wifiInf = wifiMan.getConnectionInfo();
int ipAddress = wifiInf.getIpAddress();
String ip = String.format("%d.%d.%d.%d", (ipAddress & 0xff),(ipAddress >> 8 & 0xff),(ipAddress >> 16 & 0xff),(ipAddress >> 24 & 0xff));

在Kotlin中,没有Formatter

private fun getIPAddress(useIPv4 : Boolean): String {
    try {
        var interfaces = Collections.list(NetworkInterface.getNetworkInterfaces())
        for (intf in interfaces) {
            var addrs = Collections.list(intf.getInetAddresses());
            for (addr in addrs) {
                if (!addr.isLoopbackAddress()) {
                    var sAddr = addr.getHostAddress();
                    var isIPv4: Boolean
                    isIPv4 = sAddr.indexOf(':')<0
                    if (useIPv4) {
                        if (isIPv4)
                            return sAddr;
                    } else {
                        if (!isIPv4) {
                            var delim = sAddr.indexOf('%') // drop ip6 zone suffix
                            if (delim < 0) {
                                return sAddr.toUpperCase()
                            }
                            else {
                                return sAddr.substring(0, delim).toUpperCase()
                            }
                        }
                    }
                }
            }
        }
    } catch (e: java.lang.Exception) { }
    return ""
}