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);
        }
    }
}