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));
}

给定一些字符（AllCharacters），您可以随机选择字符串中的一个字符。然后使用for循环重复获取随机字符。

public class MyProgram {
  static String getRandomString(int size) {
      String AllCharacters = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789";
      StringBuilder sb = new StringBuilder(size);
      int length = AllCharacters.length();
      for (int i = 0; i < size; i++) {
          sb.append(AllCharacters.charAt((int)(length * Math.random())));
      }
      return sb.toString();
  }

  public static void main(String[] args) {
      System.out.println(MyProgram.getRandomString(30));
  }
}

在沙盒上试试另请参阅其他语言实现随机字符串生成器

您可以创建一个包含所有字母和数字的字符数组，然后可以从该字符数组中随机选择并创建自己的字符串密码。

char[] chars = new char[62]; // Sum of letters and numbers

int i = 0;

for(char c = 'a'; c <= 'z'; c++) { // For letters
    chars[i++] = c;
}

for(char c = '0'; c <= '9';c++) { // For numbers
    chars[i++] = c;
}

for(char c = 'A'; c <= 'Z';c++) { // For capital letters
    chars[i++] = c;
}

int numberOfCodes = 0;
String code = "";
while (numberOfCodes < 1) { // Enter how much you want to generate at one time
    int numChars = 8; // Enter how many digits you want in your password

    for(i = 0; i < numChars; i++) {
        char c = chars[(int)(Math.random() * chars.length)];
        code = code + c;
    }
    System.out.println("Code is:" + code);
}

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.Random;

public class passGen{
    // Version 1.0
    private static final String dCase = "abcdefghijklmnopqrstuvwxyz";
    private static final String uCase = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    private static final String sChar = "!@#$%^&*";
    private static final String intChar = "0123456789";
    private static Random r = new Random();
    private static StringBuilder pass = new StringBuilder();

    public static void main (String[] args) {
        System.out.println ("Generating pass...");
        while (pass.length () != 16){
            int rPick = r.nextInt(4);
            if (rPick == 0){
                int spot = r.nextInt(26);
                pass.append(dCase.charAt(spot));
            } else if (rPick == 1) {
                int spot = r.nextInt(26);
                pass.append(uCase.charAt(spot));
            } else if (rPick == 2) {
                int spot = r.nextInt(8);
                pass.append(sChar.charAt(spot));
            } else {
                int spot = r.nextInt(10);
                pass.append(intChar.charAt(spot));
            }
        }
        System.out.println ("Generated Pass: " + pass.toString());
    }
}

这只是将密码添加到字符串中。。。是的，效果很好。过来看。。。这很简单；我写的。

一个简单的解决方案，但它只使用小写和数字：

Random r = new java.util.Random ();
String s = Long.toString (r.nextLong () & Long.MAX_VALUE, 36);

大小约为12位数，以36为基数，这样就无法进一步改进。当然，您可以附加多个实例。