Android设备有唯一的ID吗?如果有,使用Java访问它的简单方法是什么?
当前回答
正如DaveWebb提到的,Android开发者博客有一篇文章介绍了这一点。他们的首选解决方案是跟踪应用程序安装而不是设备,这对于大多数使用情况都很有效。博客文章将向您展示实现这一功能所需的代码,我建议您查看一下。
然而,如果您需要设备标识符而不是应用程序安装标识符,博客文章将继续讨论解决方案。我与谷歌的某位人士进行了交谈,以获得一些额外的澄清,以防您需要这样做。以下是我发现的有关设备标识符的信息,这些信息在上述博客文章中没有提及:
ANDROID_ID是首选设备标识符。ANDROID_ID在ANDROID<=2.1或>=2.3版本上非常可靠。只有2.2存在帖子中提到的问题。多个制造商的多个设备受到2.2中ANDROID_ID错误的影响。据我所知,所有受影响的设备都具有相同的ANDROID_ID,即9774d56d682e549c。顺便说一下,这也是模拟器报告的相同设备id。谷歌相信,原始设备制造商已经为他们的许多或大部分设备修补了这个问题,但我能够证实,至少在2011年4月初,找到ANDROID_ID损坏的设备仍然很容易。
根据谷歌的建议,我实现了一个类,该类将为每个设备生成一个唯一的UUID,在适当的情况下使用ANDROID_ID作为种子,必要时返回TelephonyManager.getDeviceId(),如果失败,则使用随机生成的唯一UUID,该UUID将在应用程序重新启动(但不是应用程序重新安装)期间保持。
请注意,对于必须在设备ID上回退的设备,唯一ID将在出厂重置期间保持。这是需要注意的。如果您需要确保出厂重置将重置您的唯一ID,您可能需要考虑直接返回到随机UUID而不是设备ID。
同样,此代码用于设备ID,而不是应用安装ID。对于大多数情况,应用安装ID可能是您要查找的。但是,如果您确实需要设备ID,那么下面的代码可能适用于您。
import android.content.Context;
import android.content.SharedPreferences;
import android.provider.Settings.Secure;
import android.telephony.TelephonyManager;
import java.io.UnsupportedEncodingException;
import java.util.UUID;
public class DeviceUuidFactory {
protected static final String PREFS_FILE = "device_id.xml";
protected static final String PREFS_DEVICE_ID = "device_id";
protected volatile static UUID uuid;
public DeviceUuidFactory(Context context) {
if (uuid == null) {
synchronized (DeviceUuidFactory.class) {
if (uuid == null) {
final SharedPreferences prefs = context
.getSharedPreferences(PREFS_FILE, 0);
final String id = prefs.getString(PREFS_DEVICE_ID, null);
if (id != null) {
// Use the ids previously computed and stored in the
// prefs file
uuid = UUID.fromString(id);
} else {
final String androidId = Secure.getString(
context.getContentResolver(), Secure.ANDROID_ID);
// Use the Android ID unless it's broken, in which case
// fallback on deviceId,
// unless it's not available, then fallback on a random
// number which we store to a prefs file
try {
if (!"9774d56d682e549c".equals(androidId)) {
uuid = UUID.nameUUIDFromBytes(androidId
.getBytes("utf8"));
} else {
final String deviceId = (
(TelephonyManager) context
.getSystemService(Context.TELEPHONY_SERVICE))
.getDeviceId();
uuid = deviceId != null ? UUID
.nameUUIDFromBytes(deviceId
.getBytes("utf8")) : UUID
.randomUUID();
}
} catch (UnsupportedEncodingException e) {
throw new RuntimeException(e);
}
// Write the value out to the prefs file
prefs.edit()
.putString(PREFS_DEVICE_ID, uuid.toString())
.commit();
}
}
}
}
}
/**
* Returns a unique UUID for the current android device. As with all UUIDs,
* this unique ID is "very highly likely" to be unique across all Android
* devices. Much more so than ANDROID_ID is.
*
* The UUID is generated by using ANDROID_ID as the base key if appropriate,
* falling back on TelephonyManager.getDeviceID() if ANDROID_ID is known to
* be incorrect, and finally falling back on a random UUID that's persisted
* to SharedPreferences if getDeviceID() does not return a usable value.
*
* In some rare circumstances, this ID may change. In particular, if the
* device is factory reset a new device ID may be generated. In addition, if
* a user upgrades their phone from certain buggy implementations of Android
* 2.2 to a newer, non-buggy version of Android, the device ID may change.
* Or, if a user uninstalls your app on a device that has neither a proper
* Android ID nor a Device ID, this ID may change on reinstallation.
*
* Note that if the code falls back on using TelephonyManager.getDeviceId(),
* the resulting ID will NOT change after a factory reset. Something to be
* aware of.
*
* Works around a bug in Android 2.2 for many devices when using ANDROID_ID
* directly.
*
* @see http://code.google.com/p/android/issues/detail?id=10603
*
* @return a UUID that may be used to uniquely identify your device for most
* purposes.
*/
public UUID getDeviceUuid() {
return uuid;
}
}
其他回答
检查系统信息设备唯一标识符
文档:http://docs.unity3d.com/Documentation/ScriptReference/SystemInfo-deviceUniqueIdentifier.html
唯一的设备标识符。它保证对每个设备都是唯一的(只读)。
iOS:在iOS7之前的设备上,它将返回MAC地址的哈希。在iOS7设备上,它将是UIDevice identifierForVendor,如果由于任何原因失败,则是ASIdentifierManager advertisingIdentifier。
经过多次搜索。我意识到没有确定的方法可以拥有唯一的ID。
假设我们希望每个用户只能在一部手机上使用该应用程序。
我所做的是:
当用户在我的应用程序中注册时,我将当前时间保存为服务器和应用程序数据库中的唯一ID。
当用户尝试登录另一个电话时,我从服务器获取用户信息,并意识到该用户已经登录,因为唯一ID字段已满,所以向他/她显示一个对话框,表明他/她已经登录到另一个设备,无论他/她是否想离开上一个会话,如果他说是,我将为他创建一个新的唯一ID,并更新服务器上的唯一ID详细信息。
在我自己的应用程序中,每次运行时,我都会从服务器获取用户配置文件。如果存储在服务器上的唯一ID与存储在应用程序数据库中的唯一ID不同,我将自动注销用户。
以下是我如何生成唯一id:
public static String getDeviceId(Context ctx)
{
TelephonyManager tm = (TelephonyManager) ctx.getSystemService(Context.TELEPHONY_SERVICE);
String tmDevice = tm.getDeviceId();
String androidId = Secure.getString(ctx.getContentResolver(), Secure.ANDROID_ID);
String serial = null;
if(Build.VERSION.SDK_INT > Build.VERSION_CODES.FROYO) serial = Build.SERIAL;
if(tmDevice != null) return "01" + tmDevice;
if(androidId != null) return "02" + androidId;
if(serial != null) return "03" + serial;
// other alternatives (i.e. Wi-Fi MAC, Bluetooth MAC, etc.)
return null;
}
Settings.Secure#ANDROID_ID返回每个用户64位十六进制字符串的唯一ANDROID ID。
import android.provider.Settings.Secure;
private String android_id = Secure.getString(getContext().getContentResolver(),
Secure.ANDROID_ID);
另请阅读唯一标识符的最佳实践:https://developer.android.com/training/articles/user-data-ids
我的两美分-注意,这是一个设备(错误)唯一ID,而不是Android开发者博客中讨论的安装ID。
值得注意的是,@emmby提供的解决方案在每个应用程序ID中都有所不同,因为SharedPreferences没有跨进程同步(请参阅此处和此处)。所以我完全避免了这一点。
相反,我封装了在枚举中获取(设备)ID的各种策略-更改枚举常量的顺序会影响获取ID的各种方式的优先级。返回第一个非空ID或抛出异常(根据不赋予空含义的良好Java实践)。例如,我先有一个TELEPHONY,但一个好的默认选择是ANDROID_ID贝塔:
import android.Manifest.permission;
import android.bluetooth.BluetoothAdapter;
import android.content.Context;
import android.content.pm.PackageManager;
import android.net.wifi.WifiManager;
import android.provider.Settings.Secure;
import android.telephony.TelephonyManager;
import android.util.Log;
// TODO : hash
public final class DeviceIdentifier {
private DeviceIdentifier() {}
/** @see http://code.google.com/p/android/issues/detail?id=10603 */
private static final String ANDROID_ID_BUG_MSG = "The device suffers from "
+ "the Android ID bug - its ID is the emulator ID : "
+ IDs.BUGGY_ANDROID_ID;
private static volatile String uuid; // volatile needed - see EJ item 71
// need lazy initialization to get a context
/**
* Returns a unique identifier for this device. The first (in the order the
* enums constants as defined in the IDs enum) non null identifier is
* returned or a DeviceIDException is thrown. A DeviceIDException is also
* thrown if ignoreBuggyAndroidID is false and the device has the Android ID
* bug
*
* @param ctx
* an Android constant (to retrieve system services)
* @param ignoreBuggyAndroidID
* if false, on a device with the android ID bug, the buggy
* android ID is not returned instead a DeviceIDException is
* thrown
* @return a *device* ID - null is never returned, instead a
* DeviceIDException is thrown
* @throws DeviceIDException
* if none of the enum methods manages to return a device ID
*/
public static String getDeviceIdentifier(Context ctx,
boolean ignoreBuggyAndroidID) throws DeviceIDException {
String result = uuid;
if (result == null) {
synchronized (DeviceIdentifier.class) {
result = uuid;
if (result == null) {
for (IDs id : IDs.values()) {
try {
result = uuid = id.getId(ctx);
} catch (DeviceIDNotUniqueException e) {
if (!ignoreBuggyAndroidID)
throw new DeviceIDException(e);
}
if (result != null) return result;
}
throw new DeviceIDException();
}
}
}
return result;
}
private static enum IDs {
TELEPHONY_ID {
@Override
String getId(Context ctx) {
// TODO : add a SIM based mechanism ? tm.getSimSerialNumber();
final TelephonyManager tm = (TelephonyManager) ctx
.getSystemService(Context.TELEPHONY_SERVICE);
if (tm == null) {
w("Telephony Manager not available");
return null;
}
assertPermission(ctx, permission.READ_PHONE_STATE);
return tm.getDeviceId();
}
},
ANDROID_ID {
@Override
String getId(Context ctx) throws DeviceIDException {
// no permission needed !
final String andoidId = Secure.getString(
ctx.getContentResolver(),
android.provider.Settings.Secure.ANDROID_ID);
if (BUGGY_ANDROID_ID.equals(andoidId)) {
e(ANDROID_ID_BUG_MSG);
throw new DeviceIDNotUniqueException();
}
return andoidId;
}
},
WIFI_MAC {
@Override
String getId(Context ctx) {
WifiManager wm = (WifiManager) ctx
.getSystemService(Context.WIFI_SERVICE);
if (wm == null) {
w("Wifi Manager not available");
return null;
}
assertPermission(ctx, permission.ACCESS_WIFI_STATE); // I guess
// getMacAddress() has no java doc !!!
return wm.getConnectionInfo().getMacAddress();
}
},
BLUETOOTH_MAC {
@Override
String getId(Context ctx) {
BluetoothAdapter ba = BluetoothAdapter.getDefaultAdapter();
if (ba == null) {
w("Bluetooth Adapter not available");
return null;
}
assertPermission(ctx, permission.BLUETOOTH);
return ba.getAddress();
}
}
// TODO PSEUDO_ID
// http://www.pocketmagic.net/2011/02/android-unique-device-id/
;
static final String BUGGY_ANDROID_ID = "9774d56d682e549c";
private final static String TAG = IDs.class.getSimpleName();
abstract String getId(Context ctx) throws DeviceIDException;
private static void w(String msg) {
Log.w(TAG, msg);
}
private static void e(String msg) {
Log.e(TAG, msg);
}
}
private static void assertPermission(Context ctx, String perm) {
final int checkPermission = ctx.getPackageManager().checkPermission(
perm, ctx.getPackageName());
if (checkPermission != PackageManager.PERMISSION_GRANTED) {
throw new SecurityException("Permission " + perm + " is required");
}
}
// =========================================================================
// Exceptions
// =========================================================================
public static class DeviceIDException extends Exception {
private static final long serialVersionUID = -8083699995384519417L;
private static final String NO_ANDROID_ID = "Could not retrieve a "
+ "device ID";
public DeviceIDException(Throwable throwable) {
super(NO_ANDROID_ID, throwable);
}
public DeviceIDException(String detailMessage) {
super(detailMessage);
}
public DeviceIDException() {
super(NO_ANDROID_ID);
}
}
public static final class DeviceIDNotUniqueException extends
DeviceIDException {
private static final long serialVersionUID = -8940090896069484955L;
public DeviceIDNotUniqueException() {
super(ANDROID_ID_BUG_MSG);
}
}
}
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