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

score 30 · Accepted Answer · answered Oct 07 '14 at 20:21

30

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);

answered Oct 07 '14 at 20:21

Mike Laren

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isn't Secure.ANDROID_ID supposed to be a 64 digit hash, I'm getting 16 digits. @Mike Laren – user1801879 Jun 09 '15 at 19:30
2

It's a 64-bit number than when converted to a hex string becomes 16 characters long. – Mike Laren Jun 09 '15 at 19:50
oh ok. How do I convert it to 64 bit number again, or how do I only get the bits? thanks. – user1801879 Jun 09 '15 at 19:51
2

Use `Long.parseLong(androidId, 16)` to convert it from an hex string to a `long` – Mike Laren Jun 09 '15 at 19:53
Do we need to mention any permission in manifest file for accessing android_id. – Hareesh Dec 28 '21 at 13:49
I know this is a old thread however a question, hopefully an answer.. Using targetVersion=33, same device, same line of code, however in native Android, Flutter, and Xamarin - I am getting 3 different results on the same app. Appreciate your thoughts. – Mark Anderson Jan 11 '23 at 03:21

score 3 · Answer 2 · answered Sep 16 '15 at 20:49

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;
      }
   }

score 1 · Answer 3 · answered May 10 '18 at 02:55

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

android.provider.Settings.Secure.ANDROID_ID returns "android_id"

3 Answers3