How can I get device ID for Admob

Question

I'm using Eclipse to develop applications for android, and I want to integrate Admob to make money. The tutorial says I should watch the LogCat to find ID, but where is it?

When I run in either the test mode or the real mode, sometimes the eclipse will notify that Ads returned, yet it does not show in the emu... can anyone explain?

Answer 1

The accepted answers will work if you are only testing on the Emulator or on a few devices, but if you are testing on a plethora of devices, you may need some means of prorammatically adding the running device's device ID.

The following code will make the current running device into an adview test device programmatically

...
    if(YourApplication.debugEnabled(this)) //debug flag from somewhere that you set
    {

        String android_id = Settings.Secure.getString(this.getContentResolver(), Settings.Secure.ANDROID_ID);
        String deviceId = md5(android_id).toUpperCase();
        mAdRequest.addTestDevice(deviceId);
        boolean isTestDevice = mAdRequest.isTestDevice(this);

        Log.v(TAG, "is Admob Test Device ? "+deviceId+" "+isTestDevice); //to confirm it worked
    }

You need to use the md5 of the Android ID, and it needs to be upper case. Here is the md5 code I used

public static final String md5(final String s) {
    try {
        // Create MD5 Hash
        MessageDigest digest = java.security.MessageDigest
                .getInstance("MD5");
        digest.update(s.getBytes());
        byte messageDigest[] = digest.digest();

        // Create Hex String
        StringBuffer hexString = new StringBuffer();
        for (int i = 0; i < messageDigest.length; i++) {
            String h = Integer.toHexString(0xFF & messageDigest[i]);
            while (h.length() < 2)
                h = "0" + h;
            hexString.append(h);
        }
        return hexString.toString();

    } catch (NoSuchAlgorithmException e) {
        Logger.logStackTrace(TAG,e);
    }
    return "";
}

EDIT: Apparently that MD5 method isnt perfect, and it was suggested to try https://stackoverflow.com/a/21333739/2662474 I no longer need this feature so I havent tested. Good luck!