How to find serial number of Android device?
I need to use a unique ID for an Android app and I thought the serial number for the device would be a good candidate. How do I retrieve the serial number of an Android device in my app ?
Solution 1:
TelephonyManager tManager = (TelephonyManager)myActivity.getSystemService(Context.TELEPHONY_SERVICE);
String uid = tManager.getDeviceId();
getSystemService is a method from the Activity class. getDeviceID() will return the MDN or MEID of the device depending on which radio the phone uses (GSM or CDMA).
Each device MUST return a unique value here (assuming it's a phone). This should work for any Android device with a sim slot or CDMA radio. You're on your own with that Android powered microwave ;-)
Solution 2:
As Dave Webb mentions, the Android Developer Blog has an article that covers this.
I spoke with someone at Google to get some additional clarification on a few items. Here's what I discovered that's NOT mentioned in the aforementioned blog post:
- ANDROID_ID is the preferred solution. ANDROID_ID is perfectly reliable on versions of Android <=2.1 or >=2.3. Only 2.2 has the problems mentioned in the post.
- Several devices by several manufacturers are affected by the ANDROID_ID bug in 2.2.
- 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.
- 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.
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).
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;
}
}
Solution 3:
String serial = null;
try {
Class<?> c = Class.forName("android.os.SystemProperties");
Method get = c.getMethod("get", String.class);
serial = (String) get.invoke(c, "ro.serialno");
} catch (Exception ignored) {
}
This code returns device serial number using a hidden Android API.
Solution 4:
String deviceId = Settings.System.getString(getContentResolver(),
Settings.System.ANDROID_ID);
Although, it is not guaranteed that the Android ID will be an unique identifier.
Solution 5:
There is an excellent post on the Android Developer's Blog discussing this.
It recommends against using TelephonyManager.getDeviceId() as it doesn't work on Android devices which aren't phones such as tablets, it requires the READ_PHONE_STATE permission and it doesn't work reliably on all phones.
Instead you could use one of the following:
- Mac Address
- Serial Number
- ANDROID_ID
The post discusses the pros and cons of each and it's worth reading so you can work out which would be the best for your use.