Question

我创建了一个从互联网上检索到的ArrayList个常用密码，并初始化了一个名为commonPasswords的数组。我想检查用户输入的密码是否与阵列中的任何密码匹配。但是，这似乎不起作用。我不知道为什么。任何帮助将不胜感激。

我刚开始学习如何编程，所以我在该领域相当新手。谢谢！

int commonpass = 0;
int check = 0;
while (commonpass == 0) {
    if (password.equals(commonPasswords.get(check))) {
        score = 0;
    }
    check++;
    if (check >= commonPasswords.size()) {
        commonpass++;
    }
}

Answer 1

改为使用List#contains，例如

if(commonPasswords.contains(password)){
    System.out.println("Password is not safe");
}

Answer 2

使用Java 8，您可以按照以下方式执行

List<String> commonPasswords = Arrays.asList("A", "B", "C");
return commonPasswords.stream().anyMatch(str -> str.equals(password));

Answer 3

正如钱德勒所说，你应该使用commonPasswords.contains(str)代替password.equals(commonPasswords.get(check))

if commonPasswords.contains(password)
    return true;

或者

return commonPasswords.contains(passwords);

Answer 4

因为你是初学者，所以我为你演示了以下代码，展示了4种可能非常简单的方法来做你想做的事情。建议初学者不要使用Java 8 Stream API和Lambda表达式。

    List<String> commonPasswords = Arrays.asList("touraj", "ttt", "toraj", "123");
    String userPassword = "123";

    //First Way:
    if (commonPasswords. contains(userPassword)) {
        System.out.println("Password Found");
    } else
    {
        System.out.println("Password Not Found");
    }

    //Second Way: foreach :: not suggested for beginners
    for (String commonPassword : commonPasswords) {

        if (commonPassword.equals(userPassword)) {
            System.out.println("Password Found");

            // here i use break after finding password to exit loop in order to not wasting cpu
            break;

        }

    }

    //Third Way: simple for loop :: suggested for beginners
    for (int i = 0; i <commonPasswords.size() ; i++) {

        if (commonPasswords.get(i).equals(userPassword)) {
            System.out.println("Password Found");

        }
    }

    //Forth way: Using Java 8 Stream Api :: Not Suggested for beginners like you
    boolean isPassFound =  commonPasswords.stream().anyMatch(pass -> pass.equals(userPassword));
    if (isPassFound) {
        System.out.println("Password Found.");

    }

注意：为了理解我在这里建议的java 8代码，你首先需要学习面向对象和接口，然后学习匿名方法然后学习lambda表达式然后学习Stream API ...但我认为java 8版本是自我的解释和类似于人类语言。

Answer 5

OP说他想要check to see if the user inputted password matches所以我认为他们的意思是相同的。

如果OP确实在寻找相同的条目，那么OP应该使用String.equals而不是Array.contains。因为Array.contains可能会给他一个假阳性结果。

private void checkPasswordExists(){
    List<String> passwordList = Arrays.asList("pass1", "pass2", "pass12", "pass123");

    int countContains = 0;
    String userPassword = "pass1";
    for(String password : passwordList){
        if(password.contains(userPassword)){
            countContains++;
        }
    }

    int countEquals = 0;
    for(String password : passwordList){
        if(password.equals(userPassword)){
            countEquals++;
        }
    }

    System.out.println("Password countContains = " + countContains);
    System.out.println("Password countEquals = " + countEquals);
}

以上代码写入控制台：

Password countContains = 3
Password countEquals = 1

如何检查字符串是否等于先前定义的字符串数组中的任何元素

5 个答案: