是否可以使用一些代码获得设备的IP地址?
当前回答
这里是@Nilesh和@anargund的kotlin版本
fun getIpAddress(): String {
var ip = ""
try {
val wm = applicationContext.getSystemService(WIFI_SERVICE) as WifiManager
ip = Formatter.formatIpAddress(wm.connectionInfo.ipAddress)
} catch (e: java.lang.Exception) {
}
if (ip.isEmpty()) {
try {
val en = NetworkInterface.getNetworkInterfaces()
while (en.hasMoreElements()) {
val networkInterface = en.nextElement()
val enumIpAddr = networkInterface.inetAddresses
while (enumIpAddr.hasMoreElements()) {
val inetAddress = enumIpAddr.nextElement()
if (!inetAddress.isLoopbackAddress && inetAddress is Inet4Address) {
val host = inetAddress.getHostAddress()
if (host.isNotEmpty()) {
ip = host
break;
}
}
}
}
} catch (e: java.lang.Exception) {
}
}
if (ip.isEmpty())
ip = "127.0.0.1"
return ip
}
其他回答
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?
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。
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) {}
}
}
老实说,我对代码安全只是有点熟悉,所以这可能有点像黑客。但对我来说,这是最通用的方法:
package com.my_objects.ip;
import java.net.InetAddress;
import java.net.UnknownHostException;
public class MyIpByHost
{
public static void main(String a[])
{
try
{
InetAddress host = InetAddress.getByName("nameOfDevice or webAddress");
System.out.println(host.getHostAddress());
}
catch (UnknownHostException e)
{
e.printStackTrace();
}
} }
这里是@Nilesh和@anargund的kotlin版本
fun getIpAddress(): String {
var ip = ""
try {
val wm = applicationContext.getSystemService(WIFI_SERVICE) as WifiManager
ip = Formatter.formatIpAddress(wm.connectionInfo.ipAddress)
} catch (e: java.lang.Exception) {
}
if (ip.isEmpty()) {
try {
val en = NetworkInterface.getNetworkInterfaces()
while (en.hasMoreElements()) {
val networkInterface = en.nextElement()
val enumIpAddr = networkInterface.inetAddresses
while (enumIpAddr.hasMoreElements()) {
val inetAddress = enumIpAddr.nextElement()
if (!inetAddress.isLoopbackAddress && inetAddress is Inet4Address) {
val host = inetAddress.getHostAddress()
if (host.isNotEmpty()) {
ip = host
break;
}
}
}
}
} catch (e: java.lang.Exception) {
}
}
if (ip.isEmpty())
ip = "127.0.0.1"
return ip
}
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