How to find serial number of Android device?

Question

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 devi

Answer 1

Since no answer here mentions a perfect, fail-proof ID that is both PERSISTENT through system updates and exists in ALL devices (mainly due to the fact that there isn't an individual solution from Google), I decided to post a method that is the next best thing by combining two of the available identifiers, and a check to chose between them at run-time.

Before code, 3 facts:

1. TelephonyManager.getDeviceId() (a.k.a.IMEI) will not work well or at all for non-GSM, 3G, LTE, etc. devices, but will always return a unique ID when related hardware is present, even when no SIM is inserted or even when no SIM slot exists (some OEM's have done this).

2. Since Gingerbread (Android 2.3) android.os.Build.SERIAL must exist on any device that doesn't provide IMEI, i.e., doesn't have the aforementioned hardware present, as per Android policy.

3. Due to fact (2.), at least one of these two unique identifiers will ALWAYS be present, and SERIAL can be present at the same time that IMEI is.

Note: Fact (1.) and (2.) are based on Google statements

SOLUTION

With the facts above, one can always have a unique identifier by checking if there is IMEI-bound hardware, and fall back to SERIAL when it isn't, as one cannot check if the existing SERIAL is valid. The following static class presents 2 methods for checking such presence and using either IMEI or SERIAL:

import java.lang.reflect.Method;

import android.content.Context;
import android.content.pm.PackageManager;
import android.os.Build;
import android.provider.Settings;
import android.telephony.TelephonyManager;
import android.util.Log;

public class IDManagement {

    public static String getCleartextID_SIMCHECK (Context mContext){
        String ret = "";

        TelephonyManager telMgr = (TelephonyManager) mContext.getSystemService(Context.TELEPHONY_SERVICE);

        if(isSIMAvailable(mContext,telMgr)){
            Log.i("DEVICE UNIQUE IDENTIFIER",telMgr.getDeviceId());
            return telMgr.getDeviceId();

        }
        else{
            Log.i("DEVICE UNIQUE IDENTIFIER", Settings.Secure.ANDROID_ID);

//          return Settings.Secure.ANDROID_ID;
            return android.os.Build.SERIAL;
        }
    }


    public static String getCleartextID_HARDCHECK (Context mContext){
        String ret = "";

        TelephonyManager telMgr = (TelephonyManager) mContext.getSystemService(Context.TELEPHONY_SERVICE);
        if(telMgr != null && hasTelephony(mContext)){           
            Log.i("DEVICE UNIQUE IDENTIFIER",telMgr.getDeviceId() + "");

            return telMgr.getDeviceId();    
        }
        else{
            Log.i("DEVICE UNIQUE IDENTIFIER", Settings.Secure.ANDROID_ID);

//          return Settings.Secure.ANDROID_ID;
            return android.os.Build.SERIAL;
        }
    }


    public static boolean isSIMAvailable(Context mContext, 
            TelephonyManager telMgr){

        int simState = telMgr.getSimState();

        switch (simState) {
        case TelephonyManager.SIM_STATE_ABSENT:
            return false;
        case TelephonyManager.SIM_STATE_NETWORK_LOCKED:
            return false;
        case TelephonyManager.SIM_STATE_PIN_REQUIRED:
            return false;
        case TelephonyManager.SIM_STATE_PUK_REQUIRED:
            return false;
        case TelephonyManager.SIM_STATE_READY:
            return true;
        case TelephonyManager.SIM_STATE_UNKNOWN:
            return false;
        default:
            return false;
        }
    }

    static public boolean hasTelephony(Context mContext)
    {
        TelephonyManager tm = (TelephonyManager) mContext.getSystemService(Context.TELEPHONY_SERVICE);
        if (tm == null)
            return false;

        //devices below are phones only
        if (Build.VERSION.SDK_INT < 5)
            return true;

        PackageManager pm = mContext.getPackageManager();

        if (pm == null)
            return false;

        boolean retval = false;
        try
        {
            Class [] parameters = new Class[1];
            parameters[0] = String.class;
            Method method = pm.getClass().getMethod("hasSystemFeature", parameters);
            Object [] parm = new Object[1];
            parm[0] = "android.hardware.telephony";
            Object retValue = method.invoke(pm, parm);
            if (retValue instanceof Boolean)
                retval = ((Boolean) retValue).booleanValue();
            else
                retval = false;
        }
        catch (Exception e)
        {
            retval = false;
        }

        return retval;
    }


}

I would advice on using getCleartextID_HARDCHECK. If the reflection doesn't stick in your environment, use the getCleartextID_SIMCHECK method instead, but take in consideration it should be adapted to your specific SIM-presence needs.

P.S.: Do please note that OEM's have managed to bug out SERIAL against Google policy (multiple devices with same SERIAL), and Google as stated there is at least one known case in a big OEM (not disclosed and I don't know which brand it is either, I'm guessing Samsung).

Disclaimer: This answers the original question of getting a unique device ID, but the OP introduced ambiguity by stating he needs a unique ID for an APP. Even if for such scenarios Android_ID would be better, it WILL NOT WORK after, say, a Titanium Backup of an app through 2 different ROM installs (can even be the same ROM). My solution maintains persistence that is independent of a flash or factory reset, and will only fail when IMEI or SERIAL tampering occurs through hacks/hardware mods.