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);
    }
}

public static String generateSessionKey(int length){
    String alphabet =
        new String("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"); // 9

    int n = alphabet.length(); // 10

    String result = new String();
    Random r = new Random(); // 11

    for (int i=0; i<length; i++) // 12
        result = result + alphabet.charAt(r.nextInt(n)); //13

    return result;
}

我认为这是这里最小的解决方案，或者几乎是最小的方案之一：

 public String generateRandomString(int length) {
    String randomString = "";

    final char[] chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz01234567890".toCharArray();
    final Random random = new Random();
    for (int i = 0; i < length; i++) {
        randomString = randomString + chars[random.nextInt(chars.length)];
    }

    return randomString;
}

代码工作正常。如果您正在使用此方法，我建议您使用超过10个字符。在5个字符/30362次迭代时发生冲突。这花了9秒。

我不太喜欢这些关于“简单”解决方案的答案：S

我会选择简单的；），纯Java，一行（熵基于随机字符串长度和给定字符集）：

public String randomString(int length, String characterSet) {
    return IntStream.range(0, length).map(i -> new SecureRandom().nextInt(characterSet.length())).mapToObj(randomInt -> characterSet.substring(randomInt, randomInt + 1)).collect(Collectors.joining());
}

@Test
public void buildFiveRandomStrings() {
    for (int q = 0; q < 5; q++) {
        System.out.println(randomString(10, "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")); // The character set can basically be anything
    }
}

或者（更易读的老方法）

public String randomString(int length, String characterSet) {
    StringBuilder sb = new StringBuilder(); // Consider using StringBuffer if needed
    for (int i = 0; i < length; i++) {
        int randomInt = new SecureRandom().nextInt(characterSet.length());
        sb.append(characterSet.substring(randomInt, randomInt + 1));
    }
    return sb.toString();
}

@Test
public void buildFiveRandomStrings() {
    for (int q = 0; q < 5; q++) {
        System.out.println(randomString(10, "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")); // The character set can basically be anything
    }
}

但另一方面，你也可以使用UUID，它具有相当好的熵：

UUID.randomUUID().toString().replace("-", "")

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);
}

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。