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

我开发了一个应用程序,为我的项目开发一个自动生成的字母数字字符串。在这个字符串中,前三个字符是字母,后七个字符是整数。

public class AlphaNumericGenerator {

    public static void main(String[] args) {
        java.util.Random r = new java.util.Random();
        int i = 1, n = 0;
        char c;
        String str = "";
        for (int t = 0; t < 3; t++) {
            while (true) {
                i = r.nextInt(10);
                if (i > 5 && i < 10) {

                    if (i == 9) {
                        i = 90;
                        n = 90;
                        break;
                    }
                    if (i != 90) {
                        n = i * 10 + r.nextInt(10);
                        while (n < 65) {
                            n = i * 10 + r.nextInt(10);
                        }
                    }
                    break;
                }
            }
            c = (char)n;

            str = String.valueOf(c) + str;
        }

        while(true){
            i = r.nextInt(10000000);
            if(i > 999999)
                break;
        }
        str = str + i;
        System.out.println(str);
    }
}

其他回答

我使用的是一个非常简单的Java8解决方案。只需根据您的需求进行定制。

...
import java.security.SecureRandom;
...

//Generate a random String of length between 10 to 20.
//Length is also randomly generated here.
SecureRandom random = new SecureRandom();

String sampleSet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789_";

int stringLength = random.ints(1, 10, 21).mapToObj(x -> x).reduce((a, b) -> a).get();

String randomString = random.ints(stringLength, 0, sampleSet.length() - 1)
        .mapToObj(x -> sampleSet.charAt(x))
        .collect(Collector
            .of(StringBuilder::new, StringBuilder::append,
                StringBuilder::append, StringBuilder::toString));

我们可以使用它生成如下的字母数字随机字符串(返回的字符串将强制包含一些非数字字符以及一些数字字符):

public String generateRandomString() {
            
    String sampleSet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz_";
    String sampleSetNumeric = "0123456789";
    
    String randomString = getRandomString(sampleSet, 10, 21);
    String randomStringNumeric = getRandomString(sampleSetNumeric, 10, 21);
    
    randomString = randomString + randomStringNumeric;
    
    //Convert String to List<Character>
    List<Character> list = randomString.chars()
            .mapToObj(x -> (char)x)
            .collect(Collectors.toList());
    
    Collections.shuffle(list);
    
    //This is needed to force a non-numeric character as the first String
    //Skip this for() if you don't need this logic

    for(;;) {
        if(Character.isDigit(list.get(0))) Collections.shuffle(list);
        else break;
    }
    
    //Convert List<Character> to String
    randomString = list.stream()
            .map(String::valueOf)
            .collect(Collectors.joining());
    
    return randomString;
    
}

//Generate a random number between the lower bound (inclusive) and upper bound (exclusive)
private int getRandomLength(int min, int max) {
    SecureRandom random = new SecureRandom();
    return random.ints(1, min, max).mapToObj(x -> x).reduce((a, b) -> a).get();
}

//Generate a random String from the given sample string, having a random length between the lower bound (inclusive) and upper bound (exclusive)
private String getRandomString(String sampleSet, int min, int max) {
    SecureRandom random = new SecureRandom();
    return random.ints(getRandomLength(min, max), 0, sampleSet.length() - 1)
    .mapToObj(x -> sampleSet.charAt(x))
    .collect(Collector
        .of(StringBuilder::new, StringBuilder::append,
            StringBuilder::append, StringBuilder::toString));
}

Java提供了一种直接实现这一点的方法。如果你不想要破折号,它们很容易去掉。只需使用uuid.replace(“-”,“”)

import java.util.UUID;

public class randomStringGenerator {
    public static void main(String[] args) {
        System.out.println(generateString());
    }

    public static String generateString() {
        String uuid = UUID.randomUUID().toString();
        return "uuid = " + uuid;
    }
}

输出

uuid = 2d7428a6-b58c-4008-8575-f05549f16316
import java.util.Date;
import java.util.Random;

public class RandomGenerator {

  private static Random random = new Random((new Date()).getTime());

    public static String generateRandomString(int length) {
      char[] values = {'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','0','1','2','3',
               '4','5','6','7','8','9'};

      String out = "";

      for (int i=0;i<length;i++) {
          int idx=random.nextInt(values.length);
          out += values[idx];
      }
      return out;
    }
}

使用Apache Commons库,可以在一行中完成:

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

文档

这是算盘常用的一行:

String.valueOf(CharStream.random('0', 'z').filter(c -> N.isLetterOrDigit(c)).limit(12).toArray())

随机并不意味着它必须是唯一的。要获取唯一字符串,请使用:

N.uuid() // E.g.: "e812e749-cf4c-4959-8ee1-57829a69a80f". length is 36.
N.guid() // E.g.: "0678ce04e18945559ba82ddeccaabfcd". length is 32 without '-'