Question

所以最近我一直在编写这个密码检查器，我差不多完成了。对于我的密码检查器，有一些要求。

大于8个字符
少于20个字符
必须使用@符号和下划线
还必须包含数字且不能使用名字或姓氏

我已经成功编写了所有这些，现在我只编写了一个代码，列出了所有不能同时满足的所有要求，而不是一个一个。此外，当密码可以接受并且满足所有要求时。我感谢任何帮助，谢谢大家

import java.util.Scanner;

public class PasswordChecker {
  public static void main(String[] args) {
    Scanner james = new Scanner(System.in);
    System.out.println("Hi there, Welcome to Password Heavean where we have the hardest password combinations on this easrth and any other earth.");
    System.out.println("Take note your password must be longer than 8 and less than 20 characters, Can not have your first or last name, must use a @ sign, must have a number and use at leaast 3 capitals");
    System.out.println("First please input you first name.");
    String passWord;
    String firstName;
    String lastName;
    String firstNameLower;
    String lastNameLower;
    String passWordLower;
    firstName = james.nextLine();
    firstNameLower = firstName.toLowerCase();
    System.out.println("Also please input you Last name.");
    lastName = james.nextLine();
    lastNameLower = lastName.toLowerCase();
    System.out.println("Please enter your password now");
    passWord = james.nextLine();
    passWordLower = passWord.toLowerCase();
    if (passWordLower.length() < 9) {
      while (passWordLower.length() > 20) {
        System.out.println("Sorry but your password is greater than or equal too 20 characters, please try a differnent password");
        passWord = james.nextLine();
      }
    }
    if (passWordLower.indexOf(firstNameLower) != -1 || (passWordLower.indexOf(lastNameLower) != -1)) {
      System.out.println("Sorry but you can not use your first or your last name in the password. ");
    }
    if (passWordLower.contains("@")) {} else {
      System.out.println("Sorry but you must use a @ symbol in your password.");
    }
    if (passWordLower.contains("_")) {} else {
      System.out.println("Sorry but you must use a underscore in your password.");
    }
    int counter = 0;
    for (int i = 0; i < passWordLower.length(); i++) {
      if (Character.isDigit(passWordLower.charAt(i))) {
        counter++;
      }
    }
    if (counter < 3) {
      System.out.println("Sorry but you need to have at least 3 numbers in your password.");
    }
  }
}

Answer 1

我也只用while块

更改了if + while块

while (passWord.length() < 9 || passWord.length() > 20) {
  System.out.println("Sorry but your password is greater than or equal too 20 characters, please try a differnent password");
  passWord = james.nextLine();
}
passWordLower = passWord.toLowerCase();

String errorMessage = "";
if (passWordLower.indexOf(firstNameLower) != -1 || (passWordLower.indexOf(lastNameLower) != -1)) {
  errorMessage += "Sorry but you can not use your first or your last name in the password. " + System.lineSeparator();
}
if (passWordLower.contains("@")) {} else {
  errorMessage += "Sorry but you must use a @ symbol in your password." + System.lineSeparator();
}
if (passWordLower.contains("_")) {} else {
  errorMessage += "Sorry but you must use a underscore in your password." + System.lineSeparator();
}
int counter = 0;
for (int i = 0; i < passWordLower.length(); i++) {
  if (Character.isDigit(passWordLower.charAt(i))) {
    counter++;
  }
}
if (counter < 3) {
  errorMessage += "Sorry but you need to have at least 3 numbers in your password." + System.lineSeparator();
}

System.out.println(errorMessage);

我认为你不会注意到它的区别，它打印得如此之快

Answer 2

尝试这样的事情，

 StringBuffer errorMgs = new StringBuffer();

    errorMgs.append("Sorry !! ");

    if (passWordLower.length() < 9) {
        errorMgs.append("your password must have minimum 9 characters \nand");
    }
    else if (passWordLower.length() > 20) {
        errorMgs.append("your password is greater than or equal too 20 characters \nand");
    }
    if (passWordLower.indexOf(firstNameLower) != -1 || (passWordLower.indexOf(lastNameLower) != -1)) {
        errorMgs.append(" you can not use your first or your last name \nand");
    }
    if (!passWordLower.contains("@")) {
        errorMgs.append(" you must use a @ symbol \nand");

    }
    if (!passWordLower.contains("_")) {
        errorMgs.append(" you must use a underscore \nand");
    }
    int counter = 0;
    for (int i = 0; i < passWordLower.length(); i++) {
        if (Character.isDigit(passWordLower.charAt(i))) {
            counter++;
        }
    }
    if (counter < 3) {
        errorMgs.append(" you need to have at least 3 numbers");
    }

    if (!errorMgs.toString().equals("Sorry !! ")) { //identify given password met all criterias or not   
        errorMgs.append(" in your password.");
        System.out.println(errorMgs);
    } else {
        System.out.println("your password accepted");
    }

你的结果将是这样的，

您好，欢迎来到我们最难的密码Heavean   这个easrth和任何其他地球上的密码组合。做记录   您的密码必须长于8且少于20个字符，可以   没有你的名字或姓氏，必须使用@符号，必须有   号码并在至少3个首都使用

首先请输入您的名字。

jhone

另请输入您的姓氏。

彼得

请立即输入您的密码

sld4f
     抱歉!!你必须使用@符号，你必须使用下划线和   您需要在密码中至少包含3个号码。

PassWord检查器，如何一次列出所有问题？

2 个答案: