How to Generate Unique ID in Java (Integer)?

Question

java - 如何在java中生成不猜测下一个数字的整数的唯一ID？

Answer 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.

Answer 2

int uniqueId = 0;

int getUniqueId()
{
    return uniqueId++;
}

Add synchronized if you want it to be thread safe.

Answer 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.

Answer 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.

Answer 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.

Answer 6

只需生成ID并检查它是否已存在于生成的ID列表中。

Answer 7

UUID类

Answer 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.

Answer 9

if only int is required then AtomicInteger can make it possible.

if String is needed then the below code should work by mixing timeStamp and AtomicLong.

AtomicLong idCounter =  new AtomicLong(100);
long timestamp = System.currentTimeMillis();
long nextLong = idCounter.incrementAndGet();
String randomId = String.valueOf(timestamp)+String.valueOf(nextLong); 

Answer 10

随时独特：

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