是否可以使用一些代码获得设备的IP地址?
当前回答
一个设备可能有几个IP地址,在一个特定的应用程序中使用的IP地址可能不是接收请求的服务器将看到的IP地址。事实上,一些用户使用VPN或Cloudflare Warp等代理。
如果你的目的是获得IP地址,就像服务器从你的设备接收请求一样,那么最好是通过Java客户端查询IP地理定位服务,如Ipregistry(免责声明:我为该公司工作):
https://github.com/ipregistry/ipregistry-java
IpregistryClient client = new IpregistryClient("tryout");
RequesterIpInfo requesterIpInfo = client.lookup();
requesterIpInfo.getIp();
除了使用非常简单之外,您还可以获得其他信息,例如国家、语言、货币、设备IP的时区,并且您可以识别用户是否正在使用代理。
其他回答
WifiManager wm = (WifiManager) getSystemService(WIFI_SERVICE);
String ipAddress = BigInteger.valueOf(wm.getDhcpInfo().netmask).toString();
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! !
// @NonNull
public static String getIPAddress() {
if (TextUtils.isEmpty(deviceIpAddress))
new PublicIPAddress().execute();
return deviceIpAddress;
}
public static String deviceIpAddress = "";
public static class PublicIPAddress extends AsyncTask<String, Void, String> {
InetAddress localhost = null;
protected String doInBackground(String... urls) {
try {
localhost = InetAddress.getLocalHost();
URL url_name = new URL("http://bot.whatismyipaddress.com");
BufferedReader sc = new BufferedReader(new InputStreamReader(url_name.openStream()));
deviceIpAddress = sc.readLine().trim();
} catch (Exception e) {
deviceIpAddress = "";
}
return deviceIpAddress;
}
protected void onPostExecute(String string) {
Lg.d("deviceIpAddress", string);
}
}
请检查这个代码…使用此代码。我们将从移动互联网获得IP…
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();
}
}
}
根据我的测试,这是我的建议
import java.net.*;
import java.util.*;
public class hostUtil
{
public static String HOST_NAME = null;
public static String HOST_IPADDRESS = null;
public static String getThisHostName ()
{
if (HOST_NAME == null) obtainHostInfo ();
return HOST_NAME;
}
public static String getThisIpAddress ()
{
if (HOST_IPADDRESS == null) obtainHostInfo ();
return HOST_IPADDRESS;
}
protected static void obtainHostInfo ()
{
HOST_IPADDRESS = "127.0.0.1";
HOST_NAME = "localhost";
try
{
InetAddress primera = InetAddress.getLocalHost();
String hostname = InetAddress.getLocalHost().getHostName ();
if (!primera.isLoopbackAddress () &&
!hostname.equalsIgnoreCase ("localhost") &&
primera.getHostAddress ().indexOf (':') == -1)
{
// Got it without delay!!
HOST_IPADDRESS = primera.getHostAddress ();
HOST_NAME = hostname;
//System.out.println ("First try! " + HOST_NAME + " IP " + HOST_IPADDRESS);
return;
}
for (Enumeration<NetworkInterface> netArr = NetworkInterface.getNetworkInterfaces(); netArr.hasMoreElements();)
{
NetworkInterface netInte = netArr.nextElement ();
for (Enumeration<InetAddress> addArr = netInte.getInetAddresses (); addArr.hasMoreElements ();)
{
InetAddress laAdd = addArr.nextElement ();
String ipstring = laAdd.getHostAddress ();
String hostName = laAdd.getHostName ();
if (laAdd.isLoopbackAddress()) continue;
if (hostName.equalsIgnoreCase ("localhost")) continue;
if (ipstring.indexOf (':') >= 0) continue;
HOST_IPADDRESS = ipstring;
HOST_NAME = hostName;
break;
}
}
} catch (Exception ex) {}
}
}
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