这里是Scala解决方案：

(for (i <- 0 until rnd.nextInt(64)) yield { 
  ('0' + rnd.nextInt(64)).asInstanceOf[Char] 
}) mkString("")

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

您可以使用UUID类及其getLeastSignificantBits（）消息来获取64位随机数据，然后将其转换为基数36的数字（即由0-9、a-Z组成的字符串）：

Long.toString(Math.abs( UUID.randomUUID().getLeastSignificantBits(), 36));

这将产生一个长达13个字符的字符串。我们使用Math.abs（）来确保没有负号潜入。

如果密码必须包含数字和字母特殊字符，则可以使用以下代码：

private static final String NUMBERS = "0123456789";
private static final String UPPER_ALPHABETS = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
private static final String LOWER_ALPHABETS = "abcdefghijklmnopqrstuvwxyz";
private static final String SPECIALCHARACTERS = "@#$%&*";
private static final int MINLENGTHOFPASSWORD = 8;

public static String getRandomPassword() {
    StringBuilder password = new StringBuilder();
    int j = 0;
    for (int i = 0; i < MINLENGTHOFPASSWORD; i++) {
        password.append(getRandomPasswordCharacters(j));
        j++;
        if (j == 3) {
            j = 0;
        }
    }
    return password.toString();
}

private static String getRandomPasswordCharacters(int pos) {
    Random randomNum = new Random();
    StringBuilder randomChar = new StringBuilder();
    switch (pos) {
        case 0:
            randomChar.append(NUMBERS.charAt(randomNum.nextInt(NUMBERS.length() - 1)));
            break;
        case 1:
            randomChar.append(UPPER_ALPHABETS.charAt(randomNum.nextInt(UPPER_ALPHABETS.length() - 1)));
            break;
        case 2:
            randomChar.append(SPECIALCHARACTERS.charAt(randomNum.nextInt(SPECIALCHARACTERS.length() - 1)));
            break;
        case 3:
            randomChar.append(LOWER_ALPHABETS.charAt(randomNum.nextInt(LOWER_ALPHABETS.length() - 1)));
            break;
    }
    return randomChar.toString();
}

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();
}

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);
    }
}