How to Generate Unique ID in Java (Integer)?

痞子三分冷 提交于 2019-12-17 19:17:41

问题


How to generate unique ID that is integer in java that not guess next number?


回答1:


How unique does it need to be?

If it's only unique within a process, then you can use an AtomicInteger and call incrementAndGet() each time you need a new value.




回答2:


int uniqueId = 0;

int getUniqueId()
{
    return uniqueId++;
}

Add synchronized if you want it to be thread safe.




回答3:


It's easy if you are somewhat constrained.

If you have one thread, you just use uniqueID++; Be sure to store the current uniqueID when you exit.

If you have multiple threads, a common synchronized generateUniqueID method works (Implemented the same as above).

The problem is when you have many CPUs--either in a cluster or some distributed setup like a peer-to-peer game.

In that case, you can generally combine two parts to form a single number. For instance, each process that generates a unique ID can have it's own 2-byte ID number assigned and then combine it with a uniqueID++. Something like:

return (myID << 16) & uniqueID++

It can be tricky distributing the "myID" portion, but there are some ways. You can just grab one out of a centralized database, request a unique ID from a centralized server, ...

If you had a Long instead of an Int, one of the common tricks is to take the device id (UUID) of ETH0, that's guaranteed to be unique to a server--then just add on a serial number.




回答4:


If you really meant integer rather than int:

Integer id = new Integer(42); // will not == any other Integer

If you want something visible outside a JVM to other processes or to the user, persistent, or a host of other considerations, then there are other approaches, but without context you are probably better off using using the built-in uniqueness of object identity within your system.




回答5:


 import java.util.UUID;

 public class IdGenerator {
    public static int generateUniqueId() {      
        UUID idOne = UUID.randomUUID();
        String str=""+idOne;        
        int uid=str.hashCode();
        String filterStr=""+uid;
        str=filterStr.replaceAll("-", "");
        return Integer.parseInt(str);
    }

    // XXX: replace with java.util.UUID

    public static void main(String[] args) {
        for (int i = 0; i < 5; i++) {
            System.out.println(generateUniqueId());
            //generateUniqueId();
        }
    }

}

Hope this helps you.




回答6:


Just generate ID and check whether it is already present or not in your list of generated IDs.




回答7:


UUID class




回答8:


Do you need it to be;

  • unique between two JVMs running at the same time.
  • unique even if the JVM is restarted.
  • thread-safe.
  • support null? if not, use int or long.



回答9:


Unique at any time:

int uniqueId = (int) (System.currentTimeMillis() & 0xfffffff);


来源:https://stackoverflow.com/questions/2178992/how-to-generate-unique-id-in-java-integer

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