android.provider.Settings.Secure.ANDROID_ID returns “android_id”

Question

Is there reason android.provider.Settings.Secure.ANDROID_ID is return the constant "android_id" instead of a 64-bit number as a hex string ?

Description : android.provider.Settings.Secure.ANDROID_ID

• A 64-bit number (as a hex string) that is randomly generated when the user first sets up the device and should remain constant for the lifetime of the user's device. The value may change if a factory reset is performed on the device.

I am using a Samsung Galaxy S4 w /

 <uses-sdk android:minSdkVersion="13"  android:targetSdkVersion="19" />

Cheers

Answer 1

The android.provider.Settings.Secure.ANDROID_ID is a constant that can be used in android.provider.Settings.Secure.getString(ContentResolver resolver, String name) . By default it is set to 'android_id' since that's the name of the property containing the actual Android ID.

Use this code to get the actual id:

String androidId = Secure.getString(getContext().getContentResolver(), Secure.ANDROID_ID); 

Answer 2

This is just to elaborate on the answer by @MikeLaren, which is correct. But there are a couple of gotchas to be aware of, as documented in the following code, which is what I'm currently using:

  // Get the unique (supposedly) ID for this Android device/user combination
  long androidId = convertHexToLong(Settings.Secure.getString(
                         _applicationContext.getContentResolver(), Settings.Secure.ANDROID_ID));

....

   // Method to convert a 16-character hex string into a Java long. This only took me about an hour,
   // due to a known bug in Java that it took them 13 years to fix, and still isn't fixed in the
   // version of Java used for Android development.
   // http://bugs.java.com/bugdatabase/view_bug.do?bug_id=4215269
   // http://stackoverflow.com/questions/1410168/how-to-parse-negative-long-in-hex-in-java
   // On top of that it turns out that on a HTC One the Settings.Secure.ANDROID_ID string may be 15
   // digits instead of 16!
   private long convertHexToLong(String hexString) {

      hexString = "0000000000000000" + hexString;
      int i = hexString.length();
      hexString = hexString.substring(i - 16, i);

      try {
         return Long.parseLong(hexString.substring(0, 8), 16) << 32 |
                Long.parseLong(hexString.substring(8, 16), 16);

      } catch (Exception e) {
         return 0L;
      }
   }

Answer 3

android.provider.Settings.Secure.ANDROID_ID is to big so use this answer from: https://stackoverflow.com/a/10151694/1815624

new BigInteger(string, 16).longValue()

For any value of someLong:

new BigInteger(Long.toHexString(someLong), 16).longValue() == someLong

In other words, this will return the long you sent into Long.toHexString() for any long value, including negative numbers. It will also accept strings that are bigger than a long and silently return the lower 64 bits of the string as a long. You can just check the string length <= 16 (after trimming whitespace) if you need to be sure the input fits in a long.

https://stackoverflow.com/a/10151694/1815624