如何在应用程序中获取Android手机的UUID？

I'm looking for help to get the UUID of my Android phone. I have searched the net and found one potential solution but it is not working in the emulator.

我正在寻求帮助来获取我的Android手机的UUID。我搜索了网络并找到了一个可能的解决方案，但它在模拟器中无效。

Here is the code:

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();
}

Does anybody know why it isn't working, or have a better solution?

有人知道它为什么不起作用，或者有更好的解决方案吗？

7 个解决方案

#1

This works for me:

这对我有用：

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

EDIT :

编辑：

You also need android.permission.READ_PHONE_STATE set in your Manifest. Since Android M, you need to ask this permission at runtime.

您还需要在Manifest中设置android.permission.READ_PHONE_STATE。从Android M开始，您需要在运行时询问此权限。

See this anwser : https://*.com/a/38782876/1339179

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

#2

As Dave Webb mentions, the Android Developer Blog has an article that covers this. Their preferred solution is to track app installs rather than devices, and that will work well for most use cases. The blog post will show you the necessary code to make that work, and I recommend you check it out.

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。

However, the blog post goes on to discuss solutions if you need a device identifier rather than an app installation identifier. I spoke with someone at Google to get some additional clarification on a few items in the event that you need to do so. Here's what I discovered about device identifiers that's NOT mentioned in the aforementioned blog post:

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。以下是我在上述博客文章中未提及的设备标识符的发现：

• ANDROID_ID is the preferred device identifier. ANDROID_ID is perfectly reliable on versions of Android <=2.1 or >=2.3. Only 2.2 has the problems mentioned in the post.
• 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。
• Several devices by several manufacturers are affected by the ANDROID_ID bug in 2.2.
• 几个制造商的几个设备受2.2中的ANDROID_ID错误的影响。
• As far as I've been able to determine, all affected devices have the same ANDROID_ID, which is 9774d56d682e549c. Which is also the same device id reported by the emulator, btw.
• 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。
• Google believes that OEMs have patched the issue for many or most of their devices, but I was able to verify that as of the beginning of April 2011, at least, it's still quite easy to find devices that have the broken ANDROID_ID.
• 谷歌相信原始设备制造商已经为他们的许多或大多数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。
• When a device has multiple users (available on certain devices running Android 4.2 or higher), each user appears as a completely separate device, so the ANDROID_ID value is unique to each user.
• 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。

Based on Google's recommendations, I implemented a class that will generate a unique UUID for each device, using ANDROID_ID as the seed where appropriate, falling back on TelephonyManager.getDeviceId() as necessary, and if that fails, resorting to a randomly generated unique UUID that is persisted across app restarts (but not app re-installations).

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）。

Note that for devices that have to fallback on the device ID, the unique ID WILL persist across factory resets. This is something to be aware of. If you need to ensure that a factory reset will reset your unique ID, you may want to consider falling back directly to the random UUID instead of the device ID.

请注意，对于必须回退设备ID的设备，唯一ID将在出厂重置期间保持不变。这是需要注意的事情。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。

Again, this code is for a device ID, not an app installation ID. For most situations, an app installation ID is probably what you're looking for. But if you do need a device ID, then the following code will probably work for you.

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;
    }
}

#3

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

#4

Instead of getting IMEI from TelephonyManager use ANDROID_ID.

而不是从TelephonyManager获取IMEI使用ANDROID_ID。

Settings.Secure.ANDROID_ID

This works for each android device irrespective of having telephony.

这适用于每个Android设备，无论是否具有电话功能。

#5

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

See Android Developer blog article for using UUID class to get uuid

有关使用UUID类获取uuid的信息，请参阅Android开发人员博客文章

#6

Add

加

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

Method

方法

String getUUID(){
    TelephonyManager teleManager = (TelephonyManager) getSystemService(TELEPHONY_SERVICE);
    String tmSerial = teleManager.getSimSerialNumber();
    String tmDeviceId = teleManager.getDeviceId();
    String androidId = android.provider.Settings.Secure.getString(getContentResolver(), android.provider.Settings.Secure.ANDROID_ID);
    if (tmSerial  == null) tmSerial   = "1";
    if (tmDeviceId== null) tmDeviceId = "1";
    if (androidId == null) androidId  = "1";
    UUID deviceUuid = new UUID(androidId.hashCode(), ((long)tmDeviceId.hashCode() << 32) | tmSerial.hashCode());
    String uniqueId = deviceUuid.toString();
    return uniqueId;
}

#7

As of API 26, getDeviceId() is deprecated. If you need to get the IMEI of the device, use the following:

从API 26开始，不推荐使用getDeviceId（）。如果需要获取设备的IMEI，请使用以下命令：

 String deviceId = "";
    if (Build.VERSION.SDK_INT >= 26) {
        deviceId = getSystemService(TelephonyManager.class).getImei();
    }else{
        deviceId = getSystemService(TelephonyManager.class).getDeviceId();
    }