是否可以使用某些代码获取设备的IP地址？
答案 0 :(得分：395)
这是我帮助读取IP和MAC地址的工具。实现是纯java，但我在getMACAddress()中有一个注释块，它可以读取特殊Linux（Android）文件中的值。我只在少数设备和模拟器上运行此代码，但如果您发现奇怪的结果，请告诉我。
// AndroidManifest.xml permissions
<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
// test functions
Utils.getMACAddress("wlan0");
Utils.getMACAddress("eth0");
Utils.getIPAddress(true); // IPv4
Utils.getIPAddress(false); // IPv6
<强> Utils.java
import java.io.*;
import java.net.*;
import java.util.*;
//import org.apache.http.conn.util.InetAddressUtils;
public class Utils {
/**
* Convert byte array to hex string
* @param bytes toConvert
* @return hexValue
*/
public static String bytesToHex(byte[] bytes) {
StringBuilder sbuf = new StringBuilder();
for(int idx=0; idx < bytes.length; idx++) {
int intVal = bytes[idx] & 0xff;
if (intVal < 0x10) sbuf.append("0");
sbuf.append(Integer.toHexString(intVal).toUpperCase());
}
return sbuf.toString();
}
/**
* Get utf8 byte array.
* @param str which to be converted
* @return array of NULL if error was found
*/
public static byte[] getUTF8Bytes(String str) {
try { return str.getBytes("UTF-8"); } catch (Exception ex) { return null; }
}
/**
* Load UTF8withBOM or any ansi text file.
* @param filename which to be converted to string
* @return String value of File
* @throws java.io.IOException if error occurs
*/
public static String loadFileAsString(String filename) throws java.io.IOException {
final int BUFLEN=1024;
BufferedInputStream is = new BufferedInputStream(new FileInputStream(filename), BUFLEN);
try {
ByteArrayOutputStream baos = new ByteArrayOutputStream(BUFLEN);
byte[] bytes = new byte[BUFLEN];
boolean isUTF8=false;
int read,count=0;
while((read=is.read(bytes)) != -1) {
if (count==0 && bytes[0]==(byte)0xEF && bytes[1]==(byte)0xBB && bytes[2]==(byte)0xBF ) {
isUTF8=true;
baos.write(bytes, 3, read-3); // drop UTF8 bom marker
} else {
baos.write(bytes, 0, read);
}
count+=read;
}
return isUTF8 ? new String(baos.toByteArray(), "UTF-8") : new String(baos.toByteArray());
} finally {
try{ is.close(); } catch(Exception ignored){}
}
}
/**
* Returns MAC address of the given interface name.
* @param interfaceName eth0, wlan0 or NULL=use first interface
* @return mac address or empty string
*/
public static String getMACAddress(String interfaceName) {
try {
List<NetworkInterface> interfaces = Collections.list(NetworkInterface.getNetworkInterfaces());
for (NetworkInterface intf : interfaces) {
if (interfaceName != null) {
if (!intf.getName().equalsIgnoreCase(interfaceName)) continue;
}
byte[] mac = intf.getHardwareAddress();
if (mac==null) return "";
StringBuilder buf = new StringBuilder();
for (byte aMac : mac) buf.append(String.format("%02X:",aMac));
if (buf.length()>0) buf.deleteCharAt(buf.length()-1);
return buf.toString();
}
} catch (Exception ignored) { } // for now eat exceptions
return "";
/*try {
// this is so Linux hack
return loadFileAsString("/sys/class/net/" +interfaceName + "/address").toUpperCase().trim();
} catch (IOException ex) {
return null;
}*/
}
/**
* Get IP address from first non-localhost interface
* @param useIPv4 true=return ipv4, false=return ipv6
* @return address or empty string
*/
public static String getIPAddress(boolean useIPv4) {
try {
List<NetworkInterface> interfaces = Collections.list(NetworkInterface.getNetworkInterfaces());
for (NetworkInterface intf : interfaces) {
List<InetAddress> addrs = Collections.list(intf.getInetAddresses());
for (InetAddress addr : addrs) {
if (!addr.isLoopbackAddress()) {
String sAddr = addr.getHostAddress();
//boolean isIPv4 = InetAddressUtils.isIPv4Address(sAddr);
boolean isIPv4 = sAddr.indexOf(':')<0;
if (useIPv4) {
if (isIPv4)
return sAddr;
} else {
if (!isIPv4) {
int delim = sAddr.indexOf('%'); // drop ip6 zone suffix
return delim<0 ? sAddr.toUpperCase() : sAddr.substring(0, delim).toUpperCase();
}
}
}
}
}
} catch (Exception ignored) { } // for now eat exceptions
return "";
}
}
免责声明：此Utils类的创意和示例代码来自 几个SO帖子和谷歌。我已经清理并合并了所有的例子。
答案 1 :(得分：185)
这对我有用：
WifiManager wm = (WifiManager) getSystemService(WIFI_SERVICE);
String ip = Formatter.formatIpAddress(wm.getConnectionInfo().getIpAddress());
答案 2 :(得分：60)
我使用以下代码：
我使用hashCode的原因是因为当我使用getHostAddress时，我将一些垃圾值附加到ip地址。但hashCode对我来说效果很好，因为我可以使用Formatter来获取格式正确的IP地址。
以下是示例输出：
1.使用getHostAddress：***** IP=fe80::65ca:a13d:ea5a:233d%rmnet_sdio0
2.使用hashCode和Formatter：***** IP=238.194.77.212
正如你所看到的，第二种方法完全符合我的需要。
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 ip = Formatter.formatIpAddress(inetAddress.hashCode());
Log.i(TAG, "***** IP="+ ip);
return ip;
}
}
}
} catch (SocketException ex) {
Log.e(TAG, ex.toString());
}
return null;
}
答案 3 :(得分：49)
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;
}
我添加了inetAddress instanceof Inet4Address来检查它是否是ipv4地址。
答案 4 :(得分：48)
虽然有正确的答案，但我在这里分享我的答案，并希望这种方式更方便。
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));
答案 5 :(得分：30)
下面的代码可能对您有所帮助..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" />
快乐的编码!!
答案 6 :(得分：14)
您不需要像目前为止提供的解决方案那样添加权限。以字符串形式下载此网站：
或
可以使用java代码将网站作为字符串下载：
http://www.itcuties.com/java/read-url-to-string/
解析JSON对象，如下所示：
https://stackoverflow.com/a/18998203/1987258
json属性“query”或“ip”包含IP地址。
答案 7 :(得分：9)
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;
}
答案 8 :(得分：7)
方法getDeviceIpAddress返回设备的IP地址，如果已连接，则更喜欢wifi接口地址。
return Math.abs(total);
答案 9 :(得分：3)
WifiManager wm = (WifiManager) getSystemService(WIFI_SERVICE);
String ipAddress = BigInteger.valueOf(wm.getDhcpInfo().netmask).toString();
答案 10 :(得分：2)
最近，getLocalIpAddress()仍然会返回一个IP地址，尽管与网络断开连接（没有服务指示符）。它表示设置＆gt;中显示的IP地址。关于手机＆gt;状态与应用程序的想法不同。
我之前通过添加以下代码实现了一种解决方法：
ConnectivityManager cm = getConnectivityManager();
NetworkInfo net = cm.getActiveNetworkInfo();
if ((null == net) || !net.isConnectedOrConnecting()) {
return null;
}
这对任何人都响了吗？
答案 11 :(得分：2)
这是互联网上存在的最简单的方法... 首先，将此权限添加到清单文件中...
“互联网”
“ ACCESS_NETWORK_STATE”
将此添加到Activity的onCreate文件中。
getPublicIP();
现在将此函数添加到您的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();
}
答案 12 :(得分：2)
这是this answer的重做，它去除了不相关的信息，添加了有用的注释，更清楚地命名了变量，并改善了逻辑。
别忘了包括以下权限：
<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
InternetHelper.java：
public class InternetHelper {
/**
* Get IP address from first non-localhost interface
*
* @param useIPv4 true=return ipv4, false=return ipv6
* @return address or empty string
*/
public static String getIPAddress(boolean useIPv4) {
try {
List<NetworkInterface> interfaces =
Collections.list(NetworkInterface.getNetworkInterfaces());
for (NetworkInterface interface_ : interfaces) {
for (InetAddress inetAddress :
Collections.list(interface_.getInetAddresses())) {
/* a loopback address would be something like 127.0.0.1 (the device
itself). we want to return the first non-loopback address. */
if (!inetAddress.isLoopbackAddress()) {
String ipAddr = inetAddress.getHostAddress();
boolean isIPv4 = ipAddr.indexOf(':') < 0;
if (isIPv4 && !useIPv4) {
continue;
}
if (useIPv4 && !isIPv4) {
int delim = ipAddr.indexOf('%'); // drop ip6 zone suffix
ipAddr = delim < 0 ? ipAddr.toUpperCase() :
ipAddr.substring(0, delim).toUpperCase();
}
return ipAddr;
}
}
}
} catch (Exception ignored) { } // if we can't connect, just return empty string
return "";
}
/**
* Get IPv4 address from first non-localhost interface
*
* @return address or empty string
*/
public static String getIPAddress() {
return getIPAddress(true);
}
}
答案 13 :(得分：2)
kotlin极简版本
fun getIpv4HostAddress(): String {
NetworkInterface.getNetworkInterfaces()?.toList()?.map { networkInterface ->
networkInterface.inetAddresses?.toList()?.find {
!it.isLoopbackAddress && it is Inet4Address
}?.let { return it.hostAddress }
}
return ""
}
答案 14 :(得分：1)
如果你有一个外壳; ifconfig eth0也适用于x86设备
答案 15 :(得分：1)
一台设备可能有多个IP地址，而特定应用中正在使用的IP地址可能不是接收请求的服务器将看到的IP。实际上，某些用户使用VPN或诸如Cloudflare Warp之类的代理。
如果您的目的是获取从您的设备接收请求的服务器显示的IP地址，那么最好的方法是查询IP地理位置服务，例如Ipregistry（免责声明：我为公司工作）其Java客户端：
https://github.com/ipregistry/ipregistry-java
IpregistryClient client = new IpregistryClient("tryout");
RequesterIpInfo requesterIpInfo = client.lookup();
requesterIpInfo.getIp();
除了使用起来非常简单之外，您还可以获得其他信息，例如国家/地区，语言，货币，设备IP的时区，并且可以识别用户是否正在使用代理。
答案 16 :(得分：1)
只需使用Volley从this网站获取IP
RequestQueue queue = Volley.newRequestQueue(this);
String urlip = "http://checkip.amazonaws.com/";
StringRequest stringRequest = new StringRequest(Request.Method.GET, urlip, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
txtIP.setText(response);
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
txtIP.setText("didnt work");
}
});
queue.add(stringRequest);
答案 17 :(得分：1)
这是@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
}
答案 18 :(得分：1)
在Kotlin中，没有格式化程序
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 ""
}
答案 19 :(得分：1)
在您的活动中，以下功能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();
}
}
答案 20 :(得分：0)
在更好的Kotlin解决方案中，整理一些想法以从WifiManager获取wifi ip：
private fun getWifiIp(context: Context): String? {
return context.getSystemService<WifiManager>().let {
when {
it == null -> "No wifi available"
!it.isWifiEnabled -> "Wifi is disabled"
it.connectionInfo == null -> "Wifi not connected"
else -> {
val ip = it.connectionInfo.ipAddress
((ip and 0xFF).toString() + "." + (ip shr 8 and 0xFF) + "." + (ip shr 16 and 0xFF) + "." + (ip shr 24 and 0xFF))
}
}
}
}
或者，您也可以通过NetworkInterface获取ip4环回设备的ip地址：
fun getNetworkIp4LoopbackIps(): Map<String, String> = try {
NetworkInterface.getNetworkInterfaces()
.asSequence()
.associate { it.displayName to it.ip4LoopbackIps() }
.filterValues { it.isNotEmpty() }
} catch (ex: Exception) {
emptyMap()
}
private fun NetworkInterface.ip4LoopbackIps() =
inetAddresses.asSequence()
.filter { !it.isLoopbackAddress && it is Inet4Address }
.map { it.hostAddress }
.filter { it.isNotEmpty() }
.joinToString()
答案 21 :(得分：0)
老实说，我对代码安全性只有一点点的熟悉，所以这可能有点骇人听闻。但是对我来说，这是最通用的方式：
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();
}
} }
答案 22 :(得分：0)
// @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);
}
}
答案 23 :(得分：0)
您可以做到
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);
答案 24 :(得分：0)
我不做Android，但我会以完全不同的方式解决这个问题。
向Google发送查询，例如： https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=my%20ip
并参考发布响应的HTML字段。您也可以直接查询来源。
谷歌最希望的时间超过你的应用程序。
请记住，可能是您的用户此时没有互联网，您希望发生什么！
祝你好运
答案 25 :(得分：0)
请检查此代码...使用此代码。我们将从移动互联网获取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();
}
}
}
答案 26 :(得分：-2)
根据我测试的内容，这是我的建议
ul{
max-width:87%;
background-color:grey;
display:table;
}
.floatright
{
float:right
}
li{
margin:15px;
float:left;
width: 250px;
height:270px;
background-color:pink;
display:table-cell;
margin-bottom:10px;}