Java pass 检查不起作用，条件有效

Question

import java.util.Scanner;

public class p1_l2_adrian_costin {

  /*read string from keyboard*/
  public static String read() {
      Scanner keyboard = new Scanner(System.in);
      System.out.println("Enter your password XXXXX-XXXXX-XXXXX-XXXXX \n");
      String password = keyboard.nextLine();
      return password;
  }

  /*check for number of words (4) and letters (5 per word), if ok pass true*/

  public static boolean splitAndCheckLength(String s) {
      String str[] = s.split("-");
      int counter = 0;
      for (int i = 0; i < str.length; i++)
          if (str[i].length() == 5)
              counter++;
      if (counter == 4)
          return true;
      return false;
  }

  /*check for number of words compared to numbers, if there are more numbers than words pass true
   */

  public static boolean countNumbersAndLetters(String s) {
      String str[] = s.split("-");
      int counterL = 0, counterN = 0;
      char[] c;
      for (int i = 0; i < str.length; i++) {
          c = str[i].toCharArray();
          for (int j = 0; j < c.length; j++) {
              if ((c[j] > 64 && c[j] < 91) || (c[j] > 96 && c[j] < 123))
                  counterL++;
              if (c[j] > 47 && c[j] < 58)
                  counterN++;
          }
      }
      if (counterN > counterL && counterL != 0) return true;
      return false;

  }

  public static void main(String[] args) {
      if (countNumbersAndLetters(read()) && splitAndCheckLength(read()))
          System.out.println("Good password");
      System.out.println("Bad password");
  }
}

问题是在评估每个表达式时，它们都返回 true，但是当将它们都放入 if 并尝试从主打印消息时，一切都会停止工作，并且只是再次开始输入您的密码

Answer 1

目前您调用了两次 read()，因此在调用 countNumbersAndLetters() 后，您会要求再次输入密码。

您只想获取用户输入一次:

public static void main(String[] args) {
    String input = read();
    if(countNumbersAndLetters(input) && splitAndCheckLength(input))
        System.out.println("Good password");
    else 
        System.out.println("Bad password");
}