Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

android.Settings.Secure.ANDROID_ID

I'm developing an android application and I'm going to use the LVL library to check google play licence.

After reading the LVL documentation, I've read that I have to obtain the android.Settings.Secure.ANDROID_ID to obfuscate the google play responses.

But I've also read that the ANDROID_ID obtained from TelephonyManager is sometimes the same for different devices.

First, it is not 100% reliable on releases of Android prior to 2.2 (“Froyo”). Also, there has been at least one widely-observed bug in a popular handset from a major manufacturer, where every instance has the same ANDROID_ID

Is that true?

Thanks

like image 718
Eduardo Avatar asked Jun 11 '13 14:06

Eduardo


1 Answers

Consider my case:

Same serial number on several android devices. Adb is useless. How can I change the serial number?

So I did'nt solved the problem with the ADB but to identify an Android device I use this code (use getDeviceId(context)):

public static String getDeviceId(Context context) {
    String id = getUniqueID(context);
    if (id == null)
        id = Settings.Secure.getString(context.getContentResolver(), Settings.Secure.ANDROID_ID);
    return id;
}

private static String getUniqueID(Context context) {

    String telephonyDeviceId = "NoTelephonyId";
    String androidDeviceId = "NoAndroidId";

    // get telephony id
    try {
        final TelephonyManager tm = (TelephonyManager) context.getSystemService(Context.TELEPHONY_SERVICE);
        telephonyDeviceId = tm.getDeviceId();
        if (telephonyDeviceId == null) {
            telephonyDeviceId = "NoTelephonyId";
        }
    } catch (Exception e) {
    }

    // get internal android device id
    try {
        androidDeviceId = android.provider.Settings.Secure.getString(context.getContentResolver(),
                android.provider.Settings.Secure.ANDROID_ID);
        if (androidDeviceId == null) {
            androidDeviceId = "NoAndroidId";
        }
    } catch (Exception e) {

    }

    // build up the uuid
    try {
        String id = getStringIntegerHexBlocks(androidDeviceId.hashCode())
                + "-"
                + getStringIntegerHexBlocks(telephonyDeviceId.hashCode());

        return id;
    } catch (Exception e) {
        return "0000-0000-1111-1111";
    }
}

public static String getStringIntegerHexBlocks(int value) {
    String result = "";
    String string = Integer.toHexString(value);

    int remain = 8 - string.length();
    char[] chars = new char[remain];
    Arrays.fill(chars, '0');
    string = new String(chars) + string;

    int count = 0;
    for (int i = string.length() - 1; i >= 0; i--) {
        count++;
        result = string.substring(i, i + 1) + result;
        if (count == 4) {
            result = "-" + result;
            count = 0;
        }
    }

    if (result.startsWith("-")) {
        result = result.substring(1, result.length());
    }

    return result;
}

I use it to identify a specific app installation when calling my web services. As you can see I try different approches, based also on TelephonyManager and ANDROID_ID.

What I get is a string like xxxx-xxxx-xxxx-xxxx where x is an hex character.

I bought a lot of low cost China tablets and all of these have the same DEVICE_ID and the same serial number !! So my solution. It's working well for now.

like image 95
Seraphim's Avatar answered Nov 09 '22 02:11

Seraphim's