我一直在寻找一种简单的Java算法来生成伪随机字母数字字符串。在我的情况下,它将被用作唯一的会话/密钥标识符,“很可能”在超过50万代的时间内是唯一的(我的需求实际上不需要更复杂的东西)。

理想情况下,我可以根据我的独特性需求指定长度。例如,生成的长度为12的字符串可能看起来像“AEYGF7K0DM1X”。


当前回答

static final String AB = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
static SecureRandom rnd = new SecureRandom();

String randomString(int len){
   StringBuilder sb = new StringBuilder(len);
   for(int i = 0; i < len; i++)
      sb.append(AB.charAt(rnd.nextInt(AB.length())));
   return sb.toString();
}

其他回答

public static String RandomAlphanum(int length)
{
    String charstring = "abcdefghijklmnopqrstuvwxyz0123456789";
    String randalphanum = "";
    double randroll;
    String randchar;
    for (double i = 0; i < length; i++)
    {
        randroll = Math.random();
        randchar = "";
        for (int j = 1; j <= 35; j++)
        {
            if (randroll <= (1.0 / 36.0 * j))
            {
                randchar = Character.toString(charstring.charAt(j - 1));
                break;
            }
        }
        randalphanum += randchar;
    }
    return randalphanum;
}

我使用Math.random()使用了一个非常原始的算法。为了增加随机性,可以直接实现util.Date类。尽管如此,它还是有效的。

这里是一个基于流的Java8解决方案。

    public String generateString(String alphabet, int length) {
        return generateString(alphabet, length, new SecureRandom()::nextInt);
    }

    // nextInt = bound -> n in [0, bound)
    public String generateString(String source, int length, IntFunction<Integer> nextInt) {
        StringBuilder sb = new StringBuilder();
        IntStream.generate(source::length)
                .boxed()
                .limit(length)
                .map(nextInt::apply)
                .map(source::charAt)
                .forEach(sb::append);

        return sb.toString();
    }

像这样使用

String alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
int length = 12;
String generated = generateString(alphabet, length);
System.out.println(generated);

函数nextInt应该接受一个int绑定,并返回一个介于0和绑定-1之间的随机数。

Java 8中的另一种选择是:

static final Random random = new Random(); // Or SecureRandom
static final int startChar = (int) '!';
static final int endChar = (int) '~';

static String randomString(final int maxLength) {
  final int length = random.nextInt(maxLength + 1);
  return random.ints(length, startChar, endChar + 1)
        .collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append)
        .toString();
}

使用Apache Commons库,可以在一行中完成:

import org.apache.commons.lang.RandomStringUtils;
RandomStringUtils.randomAlphanumeric(64);

文档

import java.util.*;
import javax.swing.*;

public class alphanumeric {
    public static void main(String args[]) {
        String nval, lenval;
        int n, len;

        nval = JOptionPane.showInputDialog("Enter number of codes you require: ");
        n = Integer.parseInt(nval);

        lenval = JOptionPane.showInputDialog("Enter code length you require: ");
        len = Integer.parseInt(lenval);

        find(n, len);
    }

    public static void find(int n, int length) {
        String str1 = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        StringBuilder sb = new StringBuilder(length);
        Random r = new Random();

        System.out.println("\n\t Unique codes are \n\n");
        for(int i=0; i<n; i++) {
            for(int j=0; j<length; j++) {
                sb.append(str1.charAt(r.nextInt(str1.length())));
            }
            System.out.println("  " + sb.toString());
            sb.delete(0, length);
        }
    }
}