Java isLetterOrDigit()方法、isDigit()、isLetter()

Question

我想弄清楚如何检查 String 以验证它是否至少包含一个字母和一个数字。我会提前说这是作业，我有点困惑。

有一种方法 isLetterOrDigit() 方法似乎是正确的方法，但我不确定如何在我的代码中实现它。这是我在下面使用的代码：

 import javax.swing.JOptionPane;

public class Password
{
    public static void main(String[] args)
    {

    String initialPassword;
    String secondaryPassword;
    int initialLength;

    initialPassword = JOptionPane.showInputDialog(null, "Enter Your Passowrd.");

    initialLength = initialPassword.length();

    JOptionPane.showMessageDialog(null, "initialLength = " + initialLength);

    while (initialLength < 6 || initialLength > 10)
    {
        initialPassword = JOptionPane.showInputDialog(null, "Your password does not meet the length requirements. It must be at least 6 characters long but no longer than 10.");
        initialLength = initialPassword.length();
    }

    //Needs to contain at least one letter and one digit

    secondaryPassword = JOptionPane.showInputDialog(null, "Please enter your password again to verify.");

    JOptionPane.showMessageDialog(null, "Initial password : " + initialPassword + "\nSecondar Password : " + secondaryPassword);

    while (!secondaryPassword.equals(initialPassword))
    {
        secondaryPassword = JOptionPane.showInputDialog(null, "Your passwords do not match. Please enter you password again.");
    }

    JOptionPane.showMessageDialog(null, "The program has successfully completed.");

    }
}

我想实现一种方法，其中评论部分使用 isDigit() 、 isLetter() 或 isLetterOrDigit() 方法，但我不知道怎么做它。

任何指导将不胜感激。在此先感谢您的帮助。

原文由 cdo 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

这应该工作。

 public boolean containsBothNumbersAndLetters(String password) {
  boolean digitFound = false;
  boolean letterFound = false;
  for (char ch : password.toCharArray()) {
    if (Character.isDigit(ch)) {
      digitFound = true;
    }
    if (Character.isLetter(ch)) {
      letterFound = true;
    }
    if (digitFound && letterFound) {
      // as soon as we got both a digit and a letter return true
      return true;
    }
  }
  // if not true after traversing through the entire string, return false
  return false;
}

原文由 Andreas Wederbrand 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 2

import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String str = scanner.next();
        boolean num = false, alpha = false;
        for(int index = 0; index < str.length(); index++){
          if(Character.isAlphabetic(str.charAt(index)))alpha = true;
          else if(Character.isDigit(str.charAt(index)))num = true;
        }
        System.out.print(num & alpha);
    }
}

原文由 Praveen Babu 发布，翻译遵循 CC BY-SA 4.0 许可协议