Question

我正在尝试创建一个程序，将用户的输入作为密码进行比较，以查看密码是否符合要求。如果不符合要求，请再次提示用户输入密码，直至满足要求为止。这就是我所拥有的，我不明白为什么它不起作用......

import javax.swing.*;

public class Password {

public static void main(String[] args) {
    //
    String pInput = "";

    do {
        pInput = JOptionPane.showInputDialog(null, "Please enter your password.\n"
                                                    + "Your password must have 6-10 characters\n"
                                                    + "Your password must contain at least one letter and one digit");
    }
    while (authenticate(pInput) == false);

    JOptionPane.showMessageDialog(null, "Your password was successfully entered.");
}   

private static boolean authenticate(String password)
{
    // The password should be at least six characters long.
    // The password should contain at least one letter.
    // The password should have at least one digit.
     if ((password.length() > 6) &&
        (password.length() < 10) &&
        (password.matches("[a-z]")) &&
        (password.matches("[0-9]")))
         return true;
     else
         return false;
    }
}

Answer 1

如果我理解了您的问题，我会重命名您的authenticate()方法（它真的是validate()），

// validate that a password adheres to the "rules".
private static boolean validate(String password) {
  // Check for null, then a length less then 6 (and I really don't like the length()
  // > 10 check, that's a BAD requirement).
  if (password == null || password.length() < 6 || password.length() > 10) {
    return false;
  }
  boolean containsChar = false;
  boolean containsDigit = false;
  for (char c : password.toCharArray()) {
    if (Character.isLetter(c)) {
      containsChar = true;
    } else if (Character.isDigit(c)) {
      containsDigit = true;
    }
    if (containsChar && containsDigit) {
      return true;
    }
  }
  return false;
}

就个人而言，我宁愿避免使用正则表达式，因为它们经常令人困惑。如果这是一项要求，则可以使用@MadProgrammer作为对问题的评论添加的那个，

private static final Pattern pattern = Pattern
    .compile("^(?=.*\\d)(?=.*[a-zA-Z]).{6,10}$"); // <-- note "\\d"

private static boolean validate(String password) {
  return pattern.matcher(password).matches();
}

Answer 2

你应该改变：

password.length（）＆gt; 6进入password.length（）＆gt; = 6
password.length（）＆lt; 10到password.length（）＆lt; = 10

因为你想要至少六个字符长。和大多数十个字符。
password.matches（＆＃34; [az]＆＃34;）到password.matches（＆＃34;。 [az] +。＆＃34;）因为.matches（＆＃34; [az]＆＃34;）用于检查不是字符串的字符

这个是编辑版

private static boolean authenticate(String password)
{
    // The password should be at least six characters long.
    // The password should contain at least one letter.
    // The password should have at least one digit.
     if ((password.length() >= 6) &&
        (password.length() <= 10)&&
        (password.matches(".*[a-z]+.*")) &&
        (password.matches(".*[0-9]+.*")) )
         return true;
     else
         return false;
    }
}

Answer 3

更改身份验证方法中的正则表达式：

private static boolean authenticate(String password)
{
    // The password should be at least six characters long.
    // The password should contain at least one letter.
    // The password should have at least one digit.
    if ((password.length() >= 6) &&
            (password.length() <= 10) &&
            (password.matches("^(?:.*[a-z].*)(?:.*[0-9].*)$")))
        return true;
    else
        return false;
}

使用non-capturing超前模式将启用您想要的验证。原始版本不起作用，因为它试图将密码与“[a-z]”和“[0-9]”同时匹配 - 这个条件总是返回false！

Answer 4

您需要更改：

password.length() > 6 进入： password.length() >= 6
password.length() < 10 进入： password.length() <= 10
(password.matches("[a-z]")) && (password.matches("[0-9]")) 进入： password.matches(".*[a-z]+.*") && password.matches(".*[0-9]+.*")

前两个是因为你希望6-10范围是包容性的（所以它听起来符合你的要求）第三个变化是因为您需要常规模式来匹配整个密码，而不仅仅是单个字符。

Answer 5

我认为问题在于函数的阻塞，使用以下代码

if ((password.length() >= 6) &&
        (password.length() <= 10) &&
        (password.matches(".*[a-z]+.*")) &&
        (password.matches(".*[0-9]+.*")))
         return true;

密码要求计划

5 个答案: