我想创建一个生成随机密码的程序，并要求用户猜测密码

Question

我正在创建一个程序，该程序生成随机密码，并要求用户猜测生成的密码。 我现在面临的问题是如何将生成的密码与输入密码进行比较 我的程序比较了两个密码的长度，但是我不知道比较两个密码以查看它们是否相同的方法 该程序必须比较两个密码，并且经过三次尝试，该程序必须输出生成的密码 下面的代码是我的程序

    final static Scanner in = new Scanner(System.in);
    public static String method(int len){
   String ASCII = "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 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";
        int count =0;
         int rand;
        System.out.println("Guess the generated password");
        String key = in.nextLine();
         for(int i=0;i<key.length();i++){
        count++;
    }
        SecureRandom sr = new SecureRandom();
        StringBuilder sb = new StringBuilder();
        for(int i =0;i<len;i++){
         rand= sr.nextInt(ASCII.length());
        sb.append(ASCII.charAt(rand));
        }
        for(int i=0;i<3;i++){
            if(len!=count){
            System.out.println("The length of the random generated password is "+ len+ " and the  length of the password inserted is " + count);
             System.out.println("The length of the generated password and the length of the inserted password varies. Please try again...");
             System.out.println("Guess the generated password again");
         key = in.nextLine(); 
        }
    
       }
        return sb.toString();
    
    }

 
    public static void main(String[] args) {
      

       
        Random r = new Random();

        int len;
        len = r.nextInt(1, 8);

        System.out.println("length :" + len);
        
          System.out.println(method( len));

    }

}

Answer 1

由于本文最初放置，因此您已经在编码中取得了一些进步。 基于下面显示的最新代码（从您删除的帖子中）：

final static Scanner in = new Scanner(System.in);

public static String method(int len) {
    //Create a variable to hold the ASCII character
    String ASCII = "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 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";
    int count = 0;
    int rand;
    System.out.println("Guess the generated password");
    String key = in .nextLine();
    for (int i = 0; i < key.length(); i++) {
        count++;
    }
    SecureRandom sr = new SecureRandom();
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < len; i++) {
        rand = sr.nextInt(ASCII.length());
        sb.append(ASCII.charAt(rand));
    }
    for (int i = 1; i < 3; i++) {
        while (len != key.length()) {
            System.out.println("Length do not correspond please try again");
            key = in .nextLine();
        }
        if (len == key.length() && !sb.toString().equals(key)) {
            System.out.println("Length correspond but passwords differ, Try again");
            key = in .nextLine();
        }
    }
    System.out.print("The correct password is: ");
    return sb.toString();
}


public static void main(String args[]) {
    // Create a random length generator
    Random r = new Random();
    int len;
    len = r.nextInt(1, 7);
    System.out.println("lungezza :" + len);
    System.out.println(method(len));
}

在特定范围内生成一个随机数（例如1包含7包）时，请这样做：

// int len = rand.nextInt((max - min) + 1) + min; thus...
int len = new Random().nextInt((7 - 1) + 1) + 1;

不要添加一堆ascii字符串变量中的空格。考虑根本不具备任何空格，或者如果需要的话，只需在上下字母之间提供一个：

String ASCII = "ABCDEFGHIJKLMNOPQRSTUVWXYZ abcdefghijklmnopqrstuvwxyz";

我这么说，因为可能有可能您的随机密码生成 可以创建一个密码是所有实际上不是密码的空格。

另外，我知道您的编译器不会抱怨，但是您的代码读者将...考虑使用适当的变量命名约定。即使ascii除非它将是类字段变量，否则不是。目前，它被认为是 local 变量。实际上，您在这个较旧的帖子中正确地做到了这一点。

如果尚未生成自动生成的密码，则无法猜测自动生成的密码。自动生成密码以猜测的代码是第一个要求您提供其本质的提示。

移动在上面生成密码的代码询问该密码的第一个提示的代码，例如：

//Create a variable to hold the ASCII characters
String ascii = "ABCDEFGHIJKLMNOPQRSTUVWXYZ abcdefghijklmnopqrstuvwxyz";
    
// Generate the secret password to guess...
SecureRandom sr = new SecureRandom();
StringBuilder sb = new StringBuilder("");
int rand;
for (int i = 0; i < len; i++) {
    rand = sr.nextInt(ascii.length());
    sb.append(ascii.charAt(rand));
}
    
// Guess the password...
int count = 0;
System.out.print("Guess the generated password: -> ");
String key = in .nextLine();
for (int i = 0; i < key.length(); i++) {
    count++;
} 

此 for 在第一个猜测提示下循环：

int count = 0;
System.out.print("Guess the generated password: -> ");
String key = in.nextLine();
for (int i = 0; i < key.length(); i++) {
    count++;
} 

不是必需的。它所做的就是计算第一个密码猜测中包含的字符数。具有讽刺意味的是，您已经通过仅使用key.length（）方法来知道这一点。无论如何，您都不会将Count变量用于任何内容，因此您不妨摆脱所有这些：

// Guess the password (3 guesses)...
System.out.print("Guess the generated password: -> ");
String key = in .nextLine();
for (int i = 1; i < 3; i++) {
    while (len != key.length()) {
        System.out.println("Lengths do not correspond please try again");
        key = in .nextLine();
    }
    if (len == key.length() && !sb.toString().equals(key)) {
        System.out.println("Lengths correspond but passwords differ, Try again");
        key = in .nextLine();
    }
}
System.out.print("The correct password is: ");
return sb.toString();

实际上，您只需要一个提示，并且总是很高兴验证用户的输入。这是这样做的另一种方法：

private final static Scanner in = new Scanner(System.in);
private static int numberOfGuessesAllowed = 3;
private static boolean correct = false;

public static void main(String args[]) {
    // Create a random length generator
    Random r = new Random();
    int len = r.nextInt((7 - 1) + 1) + 1;
    System.out.println("Password to guess contains " + len + " characters.");
    System.out.println("You have " + numberOfGuessesAllowed + " attempts to guess it right!");
    System.out.println(method(len));
}

public static String method(int len) {
    //Create a variable to hold the ASCII characters
    String ascii = "ABCDEFGHIJKLMNOPQRSTUVWXYZ abcdefghijklmnopqrstuvwxyz";
    
    // Generate the secret password to guess...
    SecureRandom sr = new SecureRandom();
    StringBuilder sb = new StringBuilder("");
    int rand;
    for (int i = 0; i < len; i++) {
        rand = sr.nextInt(ascii.length());
        sb.append(ascii.charAt(rand));
    }
    String secretPassword = sb.toString();
    
    
    // Guess the password...
    String key = "";
    while (key.isEmpty() && numberOfGuessesAllowed > 0) {
        System.out.print("Guess the generated password (q to quit): -> ");
        key = in .nextLine().trim();
        if (key.equalsIgnoreCase("q")) {
            return "Quitting! Bye-Bye.";
        }
        if (key.isEmpty()) {
            System.out.println("Invalid Entry! Try again...\n");
        }
        else if (secretPassword.equals(key)) {
            correct = true;
            break;
        }
        else if (len != key.length()) {
            numberOfGuessesAllowed--;
            System.out.println("Lengths do not correspond! Please try again.\n");
            key = "";
        }
        else if (len == key.length() && !secretPassword.equals(key)) {
            numberOfGuessesAllowed--;
            System.out.println("Lengths correspond but passwords differ, Try again\n");
            key = "";
        }
    }
    
    if (correct) {
        System.out.print("You guessed it! The password is: ");
    }
    else {
        System.out.print("You couldn't do it! The correct password is: ");
    }
    return secretPassword;
}