##如何獲取Android設(shè)備唯一ID？

###問題 每一個android設(shè)備都有唯一ID嗎？如果有？怎么用java最簡單取得呢？

###回答1（最佳）

如何取得android唯一碼？

好處：

• 1.不需要特定權(quán)限.
• 2.在99.5% Android裝置（包括root過的）上，即API => 9，保證唯一性.
• 3.重裝app之后仍能取得相同唯一值.

偽代碼：

if API => 9/10: (99.5% of devices)return unique ID containing serial id (rooted devices may be different)elsereturn unique ID of build information (may overlap data - API < 9)

代碼:

/*** Return pseudo unique ID* @return ID*/ @SuppressLint("NewApi") public static String getUniquePsuedoID() {// If all else fails, if the user does have lower than API 9 (lower// than Gingerbread), has reset their device 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[] supportedABIArray;if (Build.VERSION.SDK_INT >= 21) {supportedABIArray = Build.SUPPORTED_ABIS;} else {supportedABIArray = new String[] {Build.CPU_ABI};}String supportedABIs = "";try {for (String s : supportedABIArray) {supportedABIs += s;}} catch (Exception e) {supportedABIs = "";}String m_szDevIDShort = "35";if (null != Build.BOARD) m_szDevIDShort += (Build.BOARD.length() % 10);if (null != Build.BRAND) m_szDevIDShort += (Build.BRAND.length() % 10);if (null != supportedABIs) m_szDevIDShort += (supportedABIs.length() % 10);if (null != Build.DEVICE) m_szDevIDShort += (Build.DEVICE.length() % 10);if (null != Build.MANUFACTURER) m_szDevIDShort += (Build.MANUFACTURER.length() % 10);if (null != Build.MODEL) m_szDevIDShort += (Build.MODEL.length() % 10);if (null != Build.PRODUCT) m_szDevIDShort += (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 device, there will be a duplicate entryString serial = null;try {serial = android.os.Build.class.getField("SERIAL").get(null).toString();// Go ahead and return the serial for api => 9return new UUID(m_szDevIDShort.hashCode(), serial.hashCode()).toString();} catch (Exception exception) {// serial = "serial"; // constant valueserial = "" + Calendar.getInstance().getTimeInMillis(); // variable 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// identifierreturn new UUID(m_szDevIDShort.hashCode(), serial.hashCode()).toString(); }

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###回答2 好處：

• 1.不需要特定權(quán)限.
• 2.在100% Android裝置（包括root過的）上，保證唯一性.

壞處

• 1.重裝app之后不能取得相同唯一值.

private static String uniqueID = null; private static final String PREF_UNIQUE_ID = "PREF_UNIQUE_ID";public synchronized static String id(Context context) {if (uniqueID == null) {SharedPreferences sharedPrefs = context.getSharedPreferences(PREF_UNIQUE_ID, Context.MODE_PRIVATE);uniqueID = sharedPrefs.getString(PREF_UNIQUE_ID, null);if (uniqueID == null) {uniqueID = UUID.randomUUID().toString();Editor editor = sharedPrefs.edit();editor.putString(PREF_UNIQUE_ID, uniqueID);editor.commit();}}return uniqueID; }

###回答3（需要有電話卡）

好處： 1.重裝app之后仍能取得相同唯一值.

代碼：

final TelephonyManager tm = (TelephonyManager) getBaseContext().getSystemService(Context.TELEPHONY_SERVICE);final String tmDevice, tmSerial, androidId;tmDevice = "" + tm.getDeviceId();tmSerial = "" + tm.getSimSerialNumber();androidId = "" + android.provider.Settings.Secure.getString(getContentResolver(), android.provider.Settings.Secure.ANDROID_ID);UUID deviceUuid = new UUID(androidId.hashCode(), ((long)tmDevice.hashCode() << 32) | tmSerial.hashCode());String deviceId = deviceUuid.toString();

謹(jǐn)記：要取得以下權(quán)限

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

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總結(jié)

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