检查密码

Question

问题描述：

某些网站对密码规定了某些规则。编写一个检查字符串是否为有效密码的方法。假设密码规则如下：

• 密码必须至少包含八个字符。
• 密码仅包含字母和数字。
• 密码必须至少包含两位数字。

编写一个程序，提示用户输入密码，如果遵循规则则显示“有效密码”，否则显示“无效密码”。

这是我到目前为止所做的：

import java.util.*;  
import java.lang.String;  
import java.lang.Character;  

/**
 * @author CD
 * 12/2/2012
 * This class will check your password to make sure it fits the minimum set     requirements.
 */
public class CheckingPassword {  

    public static void main(String[] args) {  
        Scanner input = new Scanner(System.in);  
        System.out.print("Please enter a Password: ");  
        String password = input.next();  
        if (isValid(password)) {  
            System.out.println("Valid Password");  
        } else {  
            System.out.println("Invalid Password");  
        }  
    }  

    public static boolean isValid(String password) {  
        //return true if and only if password:
        //1. have at least eight characters.
        //2. consists of only letters and digits.
        //3. must contain at least two digits.
        if (password.length() < 8) {   
            return false;  
        } else {      
            char c;  
            int count = 1;   
            for (int i = 0; i < password.length() - 1; i++) {  
                c = password.charAt(i);  
                if (!Character.isLetterOrDigit(c)) {          
                    return false;  
                } else if (Character.isDigit(c)) {  
                    count++;  
                    if (count < 2)   {     
                        return false;  
                    }     
                }  
            }  
        }  
        return true;  
    }  
}

当我运行程序时，它只检查密码的长度，我无法弄清楚如何确保它检查字母和数字，并且密码中至少有两位数。

Answer 1

假设有效密码包含：

• 8个或更多字符，但不超过16个字符
• 一个或多个大写字符
• 一个或多个小写字符
• 一个或多个数字
• 一个或多个特殊字符（例如$，@或！）

代码：

import java.util.Scanner;
public class Password {

public static void main(String[] args) {
    // TODO Auto-generated method stub
    int min =8;
    int max=16;
    int digit=0;
    int special=0;
    int upCount=0;
    int loCount=0;
    String password;
    Scanner scan = new Scanner(System.in);
    System.out.println(" Enter Your Password:");
        password = scan.nextLine();
    if(password.length()>=min&&password.length()<=max){
        for(int i =0;i<password.length();i++){
            char c = password.charAt(i);
            if(Character.isUpperCase(c)){
                upCount++;
            }
            if(Character.isLowerCase(c)){
                loCount++;
            }
            if(Character.isDigit(c)){
                digit++;
            }
            if(c>=33&&c<=46||c==64){
                special++;
            }
        }
        if(special>=1&&loCount>=1&&upCount>=1&&digit>=1){
            System.out.println(" Password is good:");
        }

    }

    if(password.length()<min){

        for(int i =0;i<password.length();i++){
            char c = password.charAt(i);
            if(Character.isLowerCase(c)){
                loCount++;
            }
            }

        if(loCount>0){
            System.out.println(" Password must be atleat "+min+" characters:");
            System.out.println(" You need atleast one upper case chracter:");
            System.out.println(" You need atleast one digit:");
            System.out.println(" You need atleast one special chracter:");



    }
    }
    else if(password.length()<min&&upCount>1){
        for(int i =0;i<password.length();i++){
        char c =password.charAt(i);
        if(Character.isLowerCase(c)){
            loCount++;
        }
         if(Character.isUpperCase(c)){
            upCount++;
        }
        }
        if(loCount>0&&upCount>0){
        System.out.println(" Password must be atleast "+min+" chracters:");
        System.out.println(" You need atleast one digit:");
        System.out.println(" You need atleast one special chracter:");
    }
    }
     if(password.length()>max||password.length()>=max&&upCount>1&&loCount>1&&digit>1){
         System.out.println(" Password is too long.Limit is "+max+" chracters:");
                 System.out.println(" You need atleast one special chracter:");

        }
      if(password.length()>=min&&password.length()<=max&&loCount>0&&upCount>0&&digit>0&&special==0){
         System.out.println(" You need atleast a special chracter");
     }
      if(password.length()>=min&&password.length()<=max&&loCount>0&&upCount>0&&digit==0&&special==0){
         System.out.println(" You need atleast one digit:");
         System.out.println(" You need atleast one special chracter:");
     }
   }
}

Answer 2

你几乎得到了它。但是有一些错误：

• 你没有迭代密码的所有字符（i < password.length() - 1错了）
• 您的开头数字为1而不是0
• 您在检查完第一个数字后立即检查数字位数是否至少为2，而不是在扫描完所有字符后检查数字

Answer 3

如前所述，您应该首先解密所有密码字符。计算您的数字，最后检查计数是否小于2。 这是推荐代码。

if (password.length() < 8) {   
        return false;  
    } else {      
        char c;  
        int count = 0;   
        for (int i = 0; i < password.length(); i++) {  
            c = password.charAt(i);  
            if (!Character.isLetterOrDigit(c)) {          
                return false;  
            } else if (Character.isDigit(c)) {  
                count++;     
            }  
        }  
        if (count < 2)   {     
            return false;  
        }  
    }  
    return true;  
}  

Answer 4

public void run()
{
    String password= readLine("Insert Password: ");
    boolean len= true;
    boolean letter= true;
    boolean twodig= true;
       if (password.length() < 8) {   
         len = false;  
    } else {      
        char c;  
        int count = 0;   
        for (int i = 0; i < password.length(); i++) {  
            c = password.charAt(i);  
            if (!Character.isLetterOrDigit(c)) {          
                letter = false;  
            } else if (Character.isDigit(c)) {  
                count++;     
            }  
        }  
        if (count < 2)   {     
            twodig = false;  
        }  
    }  
    if(len ==true && letter == true && twodig == true)
    {
        System.out.println("This password is valid ");
    }
    else
    {
        System.out.println("This password is invalid");
    }
}

Answer 5

打包方法；

/ * 2.编写Java方法以检查字符串是否有效 密码。 密码规则： 密码必须至少包含十个字符。 密码仅包含字母和数字。 密码必须至少包含两位数字。

预期输出：

1. 密码必须至少包含八个字符。
2. 密码仅包含字母和数字。
3. 密码必须至少包含两位数字
   输入密码（您同意上述条款和条件。）：abcd1234
   密码有效：abcd1234 * /

公共类CheckPassword {

public static String password;
public static int disitCounter = 0;

public static boolean isValid(String password) {


    if (password.length() >= 10 ) {
        for(int index = 0; index < password.length(); index++) {
            char passChar = password.charAt(index);
            if (!Character.isLetterOrDigit(passChar)) {
                return false;
            }
            else {
                if (Character.isDigit(passChar)) {
                    disitCounter++;
                }
            }

        }

    }
    if(disitCounter < 2) {
        return false;
    }
    return true;
}

public static void main(String[] args) {
    password = "abcdefgh1w3";
    if(isValid(password)) {
        System.out.print("It is a valid password");
    }
    else {
        System.out.print("It is a invalid password");
    }
}

}

Answer 6

package com.parag;

/*
 * @author Parag Satav
 */
public boolean check(String password) {

    boolean flagUppercase = false;
    boolean flagLowercase = false;
    boolean flagDigit = false;
    boolean flag = false;

    if (password.length() >= 10) {

        for (int i = 0; i < password.length(); i++) {

            if (!Character.isLetterOrDigit(password.charAt(i))) {
                return false;
            }

            if (Character.isDigit(password.charAt(i)) && !flagDigit) {
                flagDigit = true;

            }

            if (Character.isUpperCase(password.charAt(i)) && !flagUppercase) {
                flagUppercase = true;

            }

            if (Character.isLowerCase(password.charAt(i)) && !flagLowercase) {
                flagLowercase = true;
            }
        }
    }

    if (flagDigit && flagUppercase && flagLowercase) {
        flag = true;
        System.out.println("Success..");
    } else
        System.out.println("Fail..");

    return flag;
}