Menu
Java's UUID class generates a random UUID. But this consists of letters and numbers. For some applications we need only numbers. Is there a way to generate random UUID that consists of only numbers in Java?
UUID.randomUUID(); 
844
Each character can be a digit 0 through 9, or letter a through f. 32 hexadecimals x log2(16) bits/hexadecimal = 128 bits in a UUID. In the version 4, variant 1 type of UUID, 6 bits are fixed and the remaining 122 bits are randomly generated, for a total of 2¹²² possible UUIDs.
From the documentation and wikipedia we see that randomUUID is good - but there is a very small chance that duplicates can be generated.
If you dont want a random number, but an UUID with numbers only use:
String lUUID = String.format("%040d", new BigInteger(UUID.randomUUID().toString().replace("-", ""), 16)); in this case left padded to 40 zeros...
results for:
UUID : b55081fa-9cd1-48c2-95d4-efe2db322a54
in:
UUID : 0241008287272164729465721528295504357972
98
For the record: UUIDs are in fact 128 bit numbers.
What you see as an alphanumeric string is the representation of that 128 bit number using hexadecimal digits (0..9A..F).
The real solution is to transform the string into it's correspondent 128 bit number. And to store that you'll need two Longs (Long has 64 bit).
26
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
