检查密码/密钥代码是否匹配的方法

Question

我想写一个简单的java程序来检查keycode是否匹配一组条件。

这是我目前所拥有的：

Scanner keycode = new Scanner(System.in);
System.out.println("Input keycode");
String key1 = keycode.nextLine();

do {

    if (key1.length() < 6 || key1.length() > 8) {

        System.out.println("must be at least 6 letter and max 8 letter");
        return;
    }

    else {

        boolean upper = false;
        boolean lower = false;
        boolean number = false;

        for (char c : key1.toCharArray()) {
            if (Character.isUpperCase(c)) {
                upper = true;
            } else if (Character.isLowerCase(c)) {
                lower = true;
            } else if (Character.isDigit(c)) {
                number = true;
            }
        }
        if (!upper) {
            System.out.println("must contain at least one uppercase character");
            return;
        } else if (!lower) {
            System.out.println("must contain at least one lowercase character");
            return;
        } else if (!number) {
            System.out.println("must contain at least one number");
            return;
        } else {
            return;
        }
    }

} while (true);

System.out.println("Input keycode again");
String key2 = keycode.nextLine();

if (key1.equals(key2)) {

    System.out.println("keycode matched");
} else {
    System.out.println("keycode dont match");

}

提示用户输入键码。

程序首先检查是大于6个字符小于8个字符。

然后检查它是否包含小写、大写和数字。

我希望它允许用户在犯任何错误时再次输入密码，而不是从头再来。

如果成功，它会要求用户再次输入键码。如果两个键码不匹配， 允许用户重试。 3次尝试失败后，系统会回复'keycode mismatch'。

如果密码不符合要求且密码不匹配，我需要帮助的部分是允许用户输入密码。当我输入少于 6 个字符的密码时，我得到以下输出：

must be at least 6 letter and max 8 letter
must be at least 6 letter and max 8 letter
must be at least 6 letter and max 8 letter
must be at least 6 letter and max 8 letter
must be at least 6 letter and max 8 letter

Answer 1

使用循环

boolean flag=false;
while(flag!=true)
{
    Scanner keycode =  new Scanner(System.in);
    System.out.println("Input keycode");
    String key1 = keycode.nextLine();
    if (key1.length() < 6 || key1.length() > 8) {

    System.out.println("must be at least 6 letter and max 8 letter");

    } 
    else
    {
         flag=true;
    }
}

Answer 2

使用boolean 变量，例如redo。并且在您使用过 return 的每个地方，将其替换为 redo = true。而你的while条件应该是while(redo)，它将一直运行到redo is true，即用户没有输入正确的数字。

另外，移动你的input阅读线：-

String key1 = keycode.nextLine();

在你的 do-while 循环中。此外，在循环中将redo 变量重置为false，这样它就不会永远保持true。

String[] keys = new String[2];  // Since you are reading 2 keys, so size = 2
boolean redo = false;
int count = 0;    // To maintain the number of keys read. should stop at count = 2

do {

    redo = false;
    Scanner keycode =  new Scanner(System.in);

    System.out.println("Input keycode No: + " + (count + 1));
    String key1 = keycode.nextLine();

    if (key1.length() < 6 || key1.length() > 8) {
        redo = true;

    } else {
        /** Your remaining code **/

        if (!upper) {
            System.out.println("must contain at least one uppercase character");
            redo = true;

        } else if (!lower) {
            System.out.println("must contain at least one lowercase character");
            redo = true;

        } else if (!number) {
            System.out.println("must contain at least one number");
            redo = true;

        } else {
            keys[count++] = key1;  // Enter a new key in array
        }
    }

} while (redo || count < 2);  

if (keys[0].equals(keys[1])) {
    System.out.println("Keys are equal");
}

Answer 3

您应该将行String key1 = keycode.nextLine(); 放在do 循环中，以便每次需要时都向用户查询一次。否则，您只是陷入了一个循环，该循环总是会一次又一次地检查相同的值。

Answer 4

boolean redo = false;

do {
    redo = false;
    Scanner keycode =  new Scanner(System.in);
    System.out.println("Input keycode");
    String key1 = keycode.nextLine();

    if (key1.length() < 6 || key1.length() > 8) {
        redo = true;

    } else {
        /** Your remaining code **/

        if (!upper) {
            System.out.println("must contain at least one uppercase character");
            redo = true;

        } else if (!lower) {
            System.out.println("must contain at least one lowercase character");
            redo = true;

        } else if (!number) {
            System.out.println("must contain at least one number");
            redo = true;
        }

          else{
          System.out.println("Input keycode again");
          String key2 = keycode.nextLine();

          if (key1.equals(key2)) {

           System.out.println("keycode matched");
            } else {
             System.out.println("keycode dont match");

             }

       }

    }

} while (redo);

我设法把代码放在里面，不知道这样做的方法是否正确？