如何在应用程序中获取Android手机的UUID？

Question

我正在寻求帮助来获取我的Android手机的UUID.我搜索了网络,发现了一个可能的解决方案,但它不能在模拟器中工作.

这是代码:

Class<?> c;
try {
    c = Class.forName("android.os.SystemProperties");
    Method get = c.getMethod("get", String.class);
    serial = (String) get.invoke(c, "ro.serialno");
    Log.d("ANDROID UUID",serial);
} catch (Exception e) {
    e.printStackTrace();
}

有人知道它为什么不起作用,或者有更好的解决方案吗？

Answer 1

正如Dave Webb所提到的,Android开发人员博客有一篇文章介绍了这一点.他们首选的解决方案是跟踪应用安装而不是设备,这对大多数用例都有效.博客文章将向您展示完成这项工作所需的代码,我建议您查看它.

但是,如果您需要设备标识符而不是应用程序安装标识符,则博客文章会继续讨论解决方案.我与Google的某位人士进行了交谈,以便在您需要的情况下对一些项目进行一些额外的澄清.以下是我在上述博文中未提及的设备标识符的发现:

• ANDROID_ID是首选的设备标识符.ANDROID_ID在Android <= 2.1或> = 2.3的版本上非常可靠.只有2.2有帖子中提到的问题.
• 几个制造商的几个设备受2.2中的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.这也是模拟器报告的相同设备ID,顺便说一下.
• 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,但我能够验证到2011年4月初,至少,找到那些破坏了ANDROID_ID的设备仍然很容易.
• 当设备有多个用户(在运行Android 4.2或更高版本的某些设备上可用)时,每个用户都显示为完全独立的设备,因此ANDROID_ID值对每个用户都是唯一的.

根据Google的建议,我实现了一个类,它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这是在应用程序重新启动时保留的(但不是应用程序重新安装).

请注意,对于必须回退设备ID的设备,唯一ID 将在出厂重置时保持不变.这是值得注意的.如果您需要确保恢复出厂设置将重置您的唯一ID,您可能需要考虑直接回退到随机UUID而不是设备ID.

同样,此代码用于设备ID,而不是应用程序安装ID.在大多数情况下,应用程序安装ID可能就是您要查找的内容.但是,如果您确实需要设备ID,那么以下代码可能适合您.

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 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;
    }
}

Answer 2

这对我有用:

TelephonyManager tManager = (TelephonyManager)getSystemService(Context.TELEPHONY_SERVICE);
String uuid = tManager.getDeviceId();

编辑:

你还需要android.permission.READ_PHONE_STATE在你的清单中设置.从Android M开始,您需要在运行时询问此权限.

请参阅此anwser:https://stackoverflow.com/a/38782876/1339179

Answer 3

<uses-permission android:name="android.permission.READ_PHONE_STATE"></uses-permission>

Answer 4

 String id = UUID.randomUUID().toString();

有关使用 UUID 类获取 uuid 的信息，请参阅Android 开发者博客文章

Answer 5

而不是从TelephonyManager获取IMEI使用ANDROID_ID.

Settings.Secure.ANDROID_ID

这适用于每个Android设备,无论是否具有电话功能.