多读书多实践,勤思考善领悟

如何获取Android设备唯一ID?

本文于1879天之前发表,文中内容可能已经过时。

问题

每一个android设备都有唯一ID吗?如果有?怎么用java最简单取得呢?

回答1(最佳)

如何取得android唯一码?

好处:

  • 1.不需要特定权限.
  • 2.在99.5% Android装置(包括root过的)上,即API => 9,保证唯一性.
  • 3.重装app之后仍能取得相同唯一值.

伪代码:

1
2
3
4
5
6
7
if API => 9/10: (99.5% of devices)

return unique ID containing serial id (rooted devices may be different)

else

return unique ID of build information (may overlap data - API < 9)

代码:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

/**
* 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 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 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 device, 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 exception) {
// 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();}

回答2

好处:

  • 1.不需要特定权限.
  • 2.在100% Android装置(包括root过的)上,保证唯一性.

坏处

  • 1.重装app之后不能取得相同唯一值.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
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(需要有电话卡)

好处:
1.重装app之后仍能取得相同唯一值.

代码:

1
2
3
4
5
6
7
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();

谨记:要取得以下权限

1
<uses-permission android:name="android.permission.READ_PHONE_STATE" />

stackoverflow链接:
http://stackoverflow.com/questions/2785485/is-there-a-unique-android-device-id