【问题标题】:Checking for repeated characters in password using Java使用Java检查密码中的重复字符
【发布时间】:2023-04-06 06:42:01
【问题描述】:

我是一名 CIS 的学生,正在学习 Java 课程(希望成为一名程序员)。无论如何,我有一个任务是创建一个检查密码有效性的程序。如果密码无效,则打印一条消息。如果密码有效,用户将重新插入密码。如果它们都匹配,则打印一条消息接受。如果它们不匹配,则打印一条消息。 密码要求是:

  • 长度必须至少为 8 个字符
    1. 必须包含一个数字
    2. 必须包含一个字母
    3. 必须包含一个特殊字符
    4. 不能有 3 个或更多相同的字符
    5. 不能包含空格
    6. 不能以 !还是?

我正在使用 netbeans。

package pass1;

import java.util.Scanner;

public class Pass1 {

//Main method asks for pasword from user and calls upon methods to verify                                 password requirements and password 1 and 2 match
public static void main(String[] args) {
    //declare variables
    String pass1;
    String pass2;
    boolean passvalid;
    boolean passmatch;


//initialize new instance of scanner
Scanner kb = new Scanner(System.in);
//get password
System.out.println("Please input password. (Must be at least 8 characters,  contain a number, letter, and special character");
pass1 = kb.nextLine();
//initialize instance of passisvalid to check password for requirements
passvalid = passisvalid(pass1);
//if pass is valid, allow user to reinsert password 
if (passvalid){
System.out.println("Please reinsert password:");
pass2 = kb.nextLine();
//initialize instance of passismatch to check passwords match
passmatch = passismatch(pass1, pass2);
//if passwords do not match, print message
if (!passmatch)
{
System.out.println("Passwords do not match.");
}
//if passwords match, print message
else if (passmatch)
{
    System.out.println("Password set.");
}
}
//if password is not valid, print message         
else
{
    System.out.println("Password is not valid:");
}


}

     /*************************************************************************************/   

//this method check that user inputted password meets requirements, and         returns boolean value 
public static boolean passisvalid(String password) {
//declare variables
boolean letter;
boolean digit;
boolean space;
boolean length;
boolean start1;
boolean start2;
boolean valid;
boolean special;
//initialize variables
valid=false;
letter=false;
digit=false;
space=false;
special=false;
length = false;
start1=false;
start2=false;

//initialize count 
for(int i=0;i<password.length();i++)
{
char s=password.charAt(i);
//check for letter in password
if(Character.isLetter(s))
{
letter = true;
}
//check for number in password
if(Character.isDigit(s))
{
digit = true;
}
//check for spaces in password
if(Character.isSpaceChar(s))
{
space=true;
}
//check for special characters in password
if(!Character.isDigit(s) && !Character.isLetter(s))
{
special=true;
}
}
//check password length
if (password.length() > 8) 
{  
length=true;
} 
//check password start with ? or !
else if (password.startsWith("?"))
{
start1=true;
}
else if (password.startsWith("!"))
{
start1=true;
}
//requirements of password for boolean true
if(letter && digit && special && length)
{
valid = true;
}
//return boolean false if detect start with ? or !, or spaces in password
if(start1||start2||space)
{
valid = false;
}

return valid;

}

/**********************************************************************************/

//this method checks that both user entered passwords match       
public static boolean passismatch(String password1, String password2){
//declare variables
boolean match;
//initialize variables
match=false;

//compare password strings   
if (password1.equals(password2))
{
match= true;
}

return match;

}

}

所以除了确保没有连续的重复字符之外,我已经能够计算出所有内容。我真的不知道从哪里开始。我在几个论坛上搜索了几个小时,试图找到一个解决方案并将它们放在一起,但没有看到一个例子,或者只是一些我一生都无法弄清楚的东西。任何帮助将不胜感激。

【问题讨论】:

  • 嗯,这听起来像是一个递归函数
  • 布尔 noConsecutive = true;从[0...n-2]循环i,比较password[i]和password[i+1],如果相同,noConsecutive = false;
  • 我不完全明白如何做到这一点,你能更深入地解释一下吗?我对循环真的很糟糕:-\。 IE。我会在哪里告诉循环转到负 2?我会用什么来比较字符?
  • 你可以看看这个问题在 SO.stackoverflow.com/questions/19605150/…
  • 是的,我见过那个。他们不要求没有连续的相同字符

标签: java validation passwords character


【解决方案1】:

您可以获取字符串数组的密码,并将每个字符串与接下来的两个字符串进行比较是否相同。

public static boolean hasConsecutiveCharacters(String pwd){
        String[] letter = pwd.split(""); // here you get each letter in to a string array

        for(int i=0; i<letter.length-2; i++){
            if(letter[i].equals(letter[i+1]) && letter[i+1].equals(letter[i+2])){
                return true; //return true as it has 3 consecutive same character
            }
        }
        return false; //If you reach here that means there are no 3 consecutive characters therefore return false.
    }

【讨论】:

  • 啊谢谢。所以使用 .split 拆分数组,将每个字符放入一个字符串中,而 letter.length-2 是我如何让它返回 2 个空格并检查字符是否相同。惊人的。我真的需要复习一下我在使用数组时遇到困难的循环和数组。谢啦。更新上面的代码,以便人们可以作为参考
【解决方案2】:
package filehandling;

import java.util.Scanner;

public class PasswordValidation {

    public static void main(String[] args) {

        String enterPassword ;
        String reEnterPassword ;

        boolean passValid ;
        boolean isPassMatch ;

        Scanner password = new Scanner(System.in);
        System.out.println("Enter Your Password:");
        System.out.println("(Must be at least 8 characters,  contain a number, letter, and special character");
        enterPassword = password.nextLine();

        passValid = checkPassIsValid(enterPassword);

        if(passValid) {
            System.out.println("Re-Enter Your Password: ");
            reEnterPassword = password.nextLine();
            isPassMatch = checkPasswordMatching(enterPassword,reEnterPassword);

            if(!isPassMatch)
                System.out.println("Password do Not Match: ");
            else if (isPassMatch)
                System.out.println("Password Set.");

        }
        else
            System.out.println("Password is Not Valid.");

    }

    private static boolean checkPassIsValid(String enterPassword) {

        boolean length = false;
        boolean letter = false;
        boolean digit = false;
        boolean space = false;
        boolean startSpace = false;
        boolean special = false;
        boolean start1 = false;
        boolean start2 = false;
        boolean valid = false;

        for(int i = 0; i < enterPassword.length(); i++) {
            char c = enterPassword.charAt(i);

            if(enterPassword.length() >= 8)
                length = true;
            if(Character.isDigit(c))
                digit = true;
            if(Character.isLetter(c))
                letter = true;
            if(Character.isSpaceChar(c))
                space = true;
            if( (!Character.isDigit(c)) && (!Character.isLetter(c)) )
                special = true;
            if(enterPassword.startsWith(" "))
                startSpace = true;
            else if(enterPassword.startsWith("?"))
                start1 = true;
            else if(enterPassword.startsWith("!"))
                start2 = true;
            if(letter && digit && special && length)
                valid = true;
            if(start1 || start2 || startSpace)
                valid = false;
        }
        return valid;

    }

    private static boolean checkPasswordMatching(String enterPassword,String reEnterPassword) {

        if(enterPassword.equals(reEnterPassword))
            return true;        
        return false;
    }

}

【讨论】:

  • 请考虑在您的帖子中添加解释。
  • 从下次开始,我记住了
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