密码检查器，如何一次列出所有问题？答案

【问题标题】：PassWord checker, How to list all problems at once?密码检查器，如何一次列出所有问题？
【发布时间】：2017-11-06 06:18:59
【问题描述】：

所以最近我一直在编写这个密码检查器，我几乎完成了。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，不能使用名字或姓氏

我已经成功地编写了所有这些代码，现在我只需编写一个代码，它会一次性列出所有未满足的要求，而不是一一列出。此外，当密码可以接受并且所有要求都已满足时。我感谢任何帮助，谢谢伙计们

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.");
    }
  }
}

【问题讨论】：

对不起，还是有点知道不知道如何正确插入看起来不错的代码。但我不认为那样看起来太糟糕了
if (passWordLower.contains("@")) {} else { 可以替换为if (!passWordLower.contains("@")) {
使用字符串缓冲区并将所有消息附加到 if 块中，最后一次全部打印出来。
谢谢，Shmosel。为我节省了一些额外且不需要的代码
我想我可以在它周围放一个大的while循环，但我不太确定。？

标签： java passwords

【解决方案1】：

试试这样的，

 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");
    }

你的输出将是这样的，

您好，欢迎来到 Password Heavean，这是我们最难的地方这个地球和任何其他地球上的密码组合。记笔记您的密码必须长于 8 个字符且少于 20 个字符，可以没有您的名字或姓氏，必须使用 @ 符号，必须有编号并使用至少 3 个大写字母

首先请输入你的名字。

jhone

另外请输入您的姓氏。

彼得

请立即输入您的密码

sld4f

对不起！！您必须使用 @ 符号并且必须使用下划线和您的密码中至少需要包含 3 个数字。

【讨论】：

只是好奇有什么办法可以摆脱 and 并换行吗？
然后使用System.lineSeparator() 而不是\n
我试过了，但是当你超过 20 个字符时它没有工作。
我觉得像这样的 di** 再问你，但如果你有 30 秒的时间不要浪费时间在上面，我怎么能让程序在密码失败后重新启动，他们可以在哪里再试一次

【解决方案2】：

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);

我不认为你会注意到差异，它打印得太快了

【讨论】：