我正在寻找在Java中创建唯一ID作为字符串的最佳方法。

任何指导都很感激,谢谢。

我应该提到我使用的是Java 5。


当前回答

恕我直言,aperkins提供了一个优雅的解决方案,因为它是原生的,使用的代码更少。 但是如果你需要一个更短的ID,你可以使用这个方法来减少生成的字符串长度:

// usage: GenerateShortUUID.next();
import java.util.UUID;

public class GenerateShortUUID() {

  private GenerateShortUUID() { } // singleton

  public static String next() {
     UUID u = UUID.randomUUID();
     return toIDString(u.getMostSignificantBits()) + toIDString(u.getLeastSignificantBits());
  }

  private static String toIDString(long i) {
      char[] buf = new char[32];
      int z = 64; // 1 << 6;
      int cp = 32;
      long b = z - 1;
      do {
          buf[--cp] = DIGITS66[(int)(i & b)];
          i >>>= 6;
      } while (i != 0);
      return new String(buf, cp, (32-cp));
  }

 // array de 64+2 digitos 
 private final static char[] DIGITS66 = {
    '0','1','2','3','4','5','6','7','8','9',        'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z',
    'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z',
    '-','.','_','~'
  };

}

其他回答

具有计数信息的唯一ID

import java.util.concurrent.atomic.AtomicLong;

public class RandomIdUtils {

    private static AtomicLong atomicCounter = new AtomicLong();

    public static String createId() {

        String currentCounter = String.valueOf(atomicCounter.getAndIncrement());
        String uniqueId = UUID.randomUUID().toString();

        return uniqueId + "-" + currentCounter;
    }
}

java.util.UUID: toString()

创建UUID。

String uniqueID = UUID.randomUUID().toString();

恕我直言,aperkins提供了一个优雅的解决方案,因为它是原生的,使用的代码更少。 但是如果你需要一个更短的ID,你可以使用这个方法来减少生成的字符串长度:

// usage: GenerateShortUUID.next();
import java.util.UUID;

public class GenerateShortUUID() {

  private GenerateShortUUID() { } // singleton

  public static String next() {
     UUID u = UUID.randomUUID();
     return toIDString(u.getMostSignificantBits()) + toIDString(u.getLeastSignificantBits());
  }

  private static String toIDString(long i) {
      char[] buf = new char[32];
      int z = 64; // 1 << 6;
      int cp = 32;
      long b = z - 1;
      do {
          buf[--cp] = DIGITS66[(int)(i & b)];
          i >>>= 6;
      } while (i != 0);
      return new String(buf, cp, (32-cp));
  }

 // array de 64+2 digitos 
 private final static char[] DIGITS66 = {
    '0','1','2','3','4','5','6','7','8','9',        'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z',
    'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z',
    '-','.','_','~'
  };

}

如果你想要简短的、人类可读的id,并且只需要它们在每次JVM运行时是唯一的:

private static long idCounter = 0;

public static synchronized String createID()
{
    return String.valueOf(idCounter++);
}    

编辑:评论中建议的替代方案-这依赖于底层的线程安全“魔法”,但更可扩展,同样安全:

private static AtomicLong idCounter = new AtomicLong();

public static String createID()
{
    return String.valueOf(idCounter.getAndIncrement());
}