Generate unique key in Java

Question

We are migrating a server project coded using NodeJs to one coded in Java. I'm not very into the cryptography thing but I need to "translate" the following instruction to Java.

crypto.randomBytes(32).toString('hex');

Basically in the node js project they were using the js library crypto to generate a unique key, and I would need to do the same, no more, no less, in java. Anyone with crypto knowledge that can help here? What would be the equivalent in Java?

Thanks

Hmm, maybe SecureRandom – Scary Wombat Aug 04 '16 at 07:03 — Scary Wombat, Aug 04 '16 at 07:03

score 2 · Answer 1 · answered Aug 04 '16 at 07:06

2

You could probably use something like this

import java.util.uuid;
...

UUID  newUUID = UUID.randomUUID();

String.valueOf(newUUID);

...

answered Aug 04 '16 at 07:06

PKey

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Deceiver · Answer 2 · 2016-08-04T07:09:25.867

1

You can use the UUID from java:

UUID.randomUUID()

Through a quick search on google I got https://paragonie.com/blog/2016/05/how-generate-secure-random-numbers-in-various-programming-languages, have a look and for your case the closest thing will be:

SecureRandom csprng = new SecureRandom();
byte[] randomBytes = new byte[32];
csprng.nextBytes(randombytes);

This is in the blog. Hope it helps.

edited Aug 04 '16 at 07:09

answered Aug 04 '16 at 06:58

Deceiver

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`would need to do the same, no more, no less` – Scary Wombat Aug 04 '16 at 06:59
ok you want actually to pass the number of bytes and generate the key – Deceiver Aug 04 '16 at 06:59
Yes, this generates a unique key. This answer is correct. – Dawood ibn Kareem Aug 04 '16 at 07:06

score 1 · Accepted Answer · edited May 23 '17 at 12:00

1

You can use

Random.nextBytes(byte[] bytes)

to populate a random byte array and then convert the bytes to hex using the strategies discussed here

edited May 23 '17 at 12:00

Community

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answered Aug 04 '16 at 07:10

John K

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`Random` doesn't provide cryptographically secure random bytes. `SecureRandom` should be used instead. – Artjom B. Aug 04 '16 at 17:23
He isn't using it for cryptography. he just needs a random UUID. – John K Aug 04 '16 at 17:39
The collisions with `Random` might be too high to call the result "unique". Anyway, [`UUID` also uses `SecureRandom` internally](http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/8u40-b25/java/util/UUID.java#UUID). – Artjom B. Aug 04 '16 at 18:58
The implementation of nextBytes is doc'd here:https://docs.oracle.com/javase/7/docs/api/java/util/Random.html#nextBytes(byte[]) so essentially it's calling nextInt() for each byte. I won't pretend to know enough about cryptography to know how much entropy this generates as opposed to other methods implemented by SecureRandom. That being said, I doubt there will be many collisions with this implementation. – John K Aug 04 '16 at 22:02

score 0 · Answer 4 · answered Aug 04 '16 at 07:15

0

Try this:

SecureRandom random = new SecureRandom();
new BigInteger(256, random).toString(32);

answered Aug 04 '16 at 07:15

Sándor Juhos

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Generate unique key in Java

4 Answers4