我正在寻求帮助以获取我的Android手机的UUID。我已经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。
以下是代码：
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？
答案 0 :(得分：98)
正如Dave Webb所提到的，Android Developer Blog has an article涵盖了这一点。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安装标识符，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：
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，则随机使用生成的唯一UUID在应用程序重新启动时保留（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 将在出厂重置期间保持不变。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，则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;
}
}
答案 1 :(得分：62)
这对我有用：
TelephonyManager tManager = (TelephonyManager)getSystemService(Context.TELEPHONY_SERVICE);
String uuid = tManager.getDeviceId();
编辑：
您还需要在清单中设置android.permission.READ_PHONE_STATE。从Android M开始，您需要在运行时询问此权限。
请参阅此anwser：https://stackoverflow.com/a/38782876/1339179
答案 2 :(得分：8)
<uses-permission android:name="android.permission.READ_PHONE_STATE"></uses-permission>
答案 3 :(得分：5)
而不是从TelephonyManager获取IMEI而不是使用ANDROID_ID。
int scoreData = whatever your number is;
LoopFunction(){
string scoreText = scoreData.toString();
string[] characters = new string[scoreText.Length];
for (int i = 0; i < text.Length; i++)
{
Instantiate (Resources.Load (characters[i])) as GameObject;
/*
Make sure that "1.pfefab" to "9.prefab" files exists in
Resources folder and 'Resources' folder exists in 'Assets' folder.
Add the offset to a variable after instantiating
so the numbers wont overlap
*/
}
}
这适用于每个Android设备，无论是否有电话。
答案 4 :(得分：2)
String id = UUID.randomUUID().toString();
请参阅Android Developer blog article for using UUID class to get uuid
答案 5 :(得分：1)
添加
<uses-permission android:name="android.permission.READ_PHONE_STATE"/>
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;
}
答案 6 :(得分：0)
从API 26开始，不推荐使用getDeviceId（）。如果您需要获取设备的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();
}