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

Question

我试图弄清楚如何检查String 以验证它是否至少包含一个字母和一个数字。我会坦率地说这是家庭作业，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."); 

    }
}

我想实现一个方法，其中评论部分使用isDigit()、isLetter() 或isLetterOrDigit() 方法，但我不知道该怎么做。

任何指导将不胜感激。提前感谢您的帮助。

Answer 1

这应该可行。

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;
}

Answer 2

如果不给你所有的代码来帮你做这件事是很难的，因为它太短了。

无论如何，首先，由于您至少需要一个字母和至少一个数字，因此您将需要两个标志，两个booleans，最初是false。您可以使用foreach 循环遍历ininitialPassword 中的每个char：

for (char c : initialPassword.toCharArray())

然后您所要做的就是在每次迭代时检查c 是否可能是字母或数字，如果是则设置相应的标志。一旦循环终止，如果两个标志都设置了，那么你的密码是有效的。这就是您的代码的样子：

boolean bHasLetter = false, bHasDigit = false;
for (char c : initialPassword.toCharArray()) {
   if (Character.isLetter(c))
      bHasLetter = true;
   else if (Character.isDigit(c))
      bHasDigit = true;

   if (bHasLetter && bHasDigit) break; // no point continuing if both found
}

if (bHasLetter && bHasDigit) { /* valid */ }

Answer 3

以下代码是我根据您的建议得出的最终代码：

import java.util.Scanner;

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

    String initialPassword;
    String secondaryPassword;
    int numLetterCheck = 0;
    int initialLength;
    boolean digitFound = false; boolean letterFound = false;


    Scanner keyboard = new Scanner(System.in);
    System.out.println("Enter a new password: ");
    initialPassword = keyboard.nextLine();

    initialLength = initialPassword.length();

    System.out.println("Your initial password length is: " + initialLength);


    while (initialLength < 6 || initialLength > 10)
    {
        System.out.println("Your password does not meet the length requirements of >6 and <10. Please enter a new password.");
        initialPassword = keyboard.nextLine();
        initialLength = initialPassword.length();
    }

    for (char ch : initialPassword.toCharArray())
    {
        if (Character.isDigit(ch))
        {
            digitFound = true;
        }
        if (Character.isLetter(ch))
        {
            letterFound = true;
        }

        if (digitFound && letterFound)
        {
            numLetterCheck = 0;
        }
        else
        {
            numLetterCheck = 1;
        }
    } 

    while (numLetterCheck == 1)
    {
        System.out.println("Your password must contain at least one number and one number. Please enter a new passord that meets this criteria: ");
        initialPassword = keyboard.nextLine();

        for (char ch : initialPassword.toCharArray())
        {
            if (Character.isDigit(ch))
            {
                digitFound = true;
            }
            if (Character.isLetter(ch))
            {
                letterFound = true;
            }

            if (digitFound && letterFound)
            {
                numLetterCheck = 0;
            }
            else
            {
                numLetterCheck = 1;
            }
        }
    }

    System.out.println("Please enter your password again to verify it's accuracy; ");
    secondaryPassword = keyboard.nextLine();

    System.out.println("Initial password : " + initialPassword + "\nSecondar Password : " + secondaryPassword);

   while (!secondaryPassword.equals(initialPassword))
{
    System.out.println("Your passwords do not match. Please enter your password again to verify.");
    secondaryPassword = keyboard.nextLine();    
}

System.out.println("The program has successfully completed.");  

}

}

Answer 4

这似乎是一个老问题，之前已经回答过了，但我正在添加我的代码，因为我遇到了泰语重音字符的问题。所以我努力解决这个问题，我发现上面的解决方案，如果你正在处理这样的字符，这是不完整的 - ก่อนที่สุด ท้ายo

为了正确识别这些字符，代码如下：

String value = "abc123ก่อนที่สุด ท้ายo";
    // Loop through characters in this String.
    for (int i = 0; i < value.length(); i++) {
        char c = value.charAt(i);

        // See if the character is a letter or not.
        if (Character.isLetter(c)) {
        System.out.println("This " + c + " = LETTER");
        } 
        if (Character.isDigit(c)) {
        System.out.println("This " + c + " DIGIT");
        }

        if ((""+c).matches("\\p{M}"))
            System.out.println("This " + c + " = UNICODE LETTER");
    }

不确定是否有人也遇到过这种情况。希望这会有所帮助。

Answer 5

您可以使用以下代码：

import java.util.Scanner;

public class IsDigitisLetter {

    public static void main(String[] args) {
        Scanner scnr = new Scanner(System.in);

        char character;
        System.out.println("Please enter a single character: ");
        character = scnr.next().charAt(0);

        System.out.println(character);

        if (Character.isLetter(character)) {
            System.out.println("The character entered is a letter.");
        } else if (Character.isDigit(character)) {
            System.out.println("The character entered is a digit.");
        }
    }

}

Answer 6

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);
    }
}