博客摘要：很简单，就是获取设备的唯一ID，直接上正文。
先来看看几种比较单一的方式：
IMEI
方式：TelephonyManager.getDeviceId():
问题
范围：只能支持拥有通话功能的设备，对于平板不可以。
持久性：返厂，数据擦除的时候不彻底，保留了原来的标识。
权限：需要权限：android.permission.READ_PHONE_STATE
bug: 有些厂家的实现有bug，返回一些不可用的数据
Mac地址
ACCESS_WIFI_STATE权限
有些设备没有WiFi，或者蓝牙，就不可以，如果WiFi没有打开，硬件也不会返回Mac地址，不建议使用
ANDROID_ID
2.2（Froyo，8）版本系统会不可信，来自主要生产厂商的主流手机，至少有一个普遍发现的bug，这些有问题的手机相同的ANDROID_ID: 9774d56d682e549c
但是如果返厂的手机，或者被root的手机，可能会变
Serial Number
从Android 2.3 (“Gingerbread”)开始可用，可以通过android.os.Build.SERIAL获取，对于没有通话功能的设备，它会
返回一个唯一的device ID，
以下几个是stackoverflow上评论较多的几个，没贴完，还有其他，综合的，用到以上的部分方式：
地址：http://stackoverflow.com/questions/2785485/is-there-a-unique-android-device-id
google官方的相关博客：http://android-developers.blogspot.com/2011/03/identifying-app-installations.html
有兴趣的朋友可以再仔细看看
支持率比较高的（支持票数157）：androidID --> 剔除2.2版本（API 8）中有问题的手机，使用UUID替代
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 volatile 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;
}
}
根据版本进行判断的方式：Serial序列号-->UUID （支持数31）
通过Serial 即可，在覆盖率上，你已经成功的获得了98.4%的用户，剩下的1.6%的用户系统是在9 以下的。
通过AndroidID获取，前面已经说过，在8上，有些商家的手机会有一些bug，返回相同的AndroidID，如果Serial和AndroidID都不行
/**
Return pseudo unique ID
-
@return ID
*/
public static String getUniquePsuedoID()
{
// If all else fails, if the user does have lower than API 9 (lower
// than Gingerbread), has reset their phone or 'Secure.ANDROID_ID'
// returns 'null', then simply the ID returned will be solely based
// off their Android device information. This is where the collisions
// can happen.
// Thanks http://www.pocketmagic.net/?p=1662!
// Try not to use DISPLAY, HOST or ID - these items could change.
// If there are collisions, there will be overlapping data
String m_szDevIDShort = "35" + (Build.BOARD.length() % 10) + (Build.BRAND.length() % 10) + (Build.CPU_ABI.length() % 10) + (Build.DEVICE.length() % 10) + (Build.MANUFACTURER.length() % 10) + (Build.MODEL.length() % 10) + (Build.PRODUCT.length() % 10);// Thanks to @Roman SL!
// http://stackoverflow.com/a/4789483/950427
// Only devices with API >= 9 have android.os.Build.SERIAL
// http://developer.android.com/reference/android/os/Build.html#SERIAL
// If a user upgrades software or roots their phone, there will be a duplicate entry
String serial = null;
try
{
serial = android.os.Build.class.getField("SERIAL").get(null).toString();// Go ahead and return the serial for api => 9 return new UUID(m_szDevIDShort.hashCode(), serial.hashCode()).toString();}
catch (Exception e)
{
// String needs to be initialized
serial = "serial"; // some value
}// Thanks @Joe!
// http://stackoverflow.com/a/2853253/950427
// Finally, combine the values we have found by using the UUID class to create a unique identifier
return new UUID(m_szDevIDShort.hashCode(), serial.hashCode()).toString();
}
不用READ_PHONE_STATE权限直接获取ROM信息的方式：（支持率较低 16）
String m_szDevIDShort = "35" + //we make this look like a valid IMEI
Build.BOARD.length()%10+ Build.BRAND.length()%10 +
Build.CPU_ABI.length()%10 + Build.DEVICE.length()%10 +
Build.DISPLAY.length()%10 + Build.HOST.length()%10 +
Build.ID.length()%10 + Build.MANUFACTURER.length()%10 +
Build.MODEL.length()%10 + Build.PRODUCT.length()%10 +
Build.TAGS.length()%10 + Build.TYPE.length()%10 +
Build.USER.length()%10 ; //13 digits
最后贴上自己在项目中用的：
public static String getDeviceId(Context context) {
String deviceId = "";
if (deviceId != null && !"".equals(deviceId)) {
return deviceId;
}
if (deviceId == null || "".equals(deviceId)) {
try {
deviceId = getLocalMac(context).replace(":", "");
} catch (Exception e) {
e.printStackTrace();
}
}
if (deviceId == null || "".equals(deviceId)) {
try {
deviceId = getAndroidId(context);
} catch (Exception e) {
e.printStackTrace();
}
}
if (deviceId == null || "".equals(deviceId)) {
if (deviceId == null || "".equals(deviceId)) {
UUID uuid = UUID.randomUUID();
deviceId = uuid.toString().replace("-", "");
writeDeviceID(deviceId);
}
}
return deviceId;
}
// IMEI码
private static String getIMIEStatus(Context context) {
TelephonyManager tm = (TelephonyManager) context
.getSystemService(Context.TELEPHONY_SERVICE);
String deviceId = tm.getDeviceId();
return deviceId;
}
// Mac地址
private static String getLocalMac(Context context) {
WifiManager wifi = (WifiManager) context
.getSystemService(Context.WIFI_SERVICE);
WifiInfo info = wifi.getConnectionInfo();
return info.getMacAddress();
}
// Android Id
private static String getAndroidId(Context context) {
String androidId = Settings.Secure.getString(
context.getContentResolver(), Settings.Secure.ANDROID_ID);
return androidId;
}
public static void saveDeviceID(String str) {
try {
FileOutputStream fos = new FileOutputStream(file);
Writer out = new OutputStreamWriter(fos, "UTF-8");
out.write(str);
out.close();
} catch (IOException e) {
e.printStackTrace();
}
}
public static String readDeviceID() {
StringBuffer buffer = new StringBuffer();
try {
FileInputStream fis = new FileInputStream(file);
InputStreamReader isr = new InputStreamReader(fis, "UTF-8");
Reader in = new BufferedReader(isr);
int i;
while ((i = in.read()) > -1) {
buffer.append((char) i);
}
in.close();
return buffer.toString();
} catch (IOException e) {
e.printStackTrace();
return null;
}
}
对于获取设备唯一ID并没有绝对的方案，这一点在android的官方博客中也提到了，不过以上几种方案，应该可以满足平时的需求，大家可以选择其中自己认为比较好的，用于自己的项目中。不知道其他朋友在项目中是如何处理的，欢迎交流讨论。