Question

<?php

class Validator {

public $errors = array(
        'password' => '', 
        'email' => '');

const PASSWORD_MINCHARS = 8;

public function checkEmail($email) {

    if ($this->checkEmpty($email)) {
        if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
            $this->errors['email'] = "Please provide a valid email";
            return FALSE;
        } else {
            return TRUE;
        }   
    } else {
        $this->errors['email'] = "Please provide a value for the email";
        return FALSE;
    }
}

public function checkPassword($string) {

    if ($this->checkEmpty($string)) {

        if (strlen($string) < self::PASSWORD_MINCHARS) {
            $this->errors['password'] = "The password should be atleast ".self::PASSWORD_MINCHARS." characters long.";
            return FALSE;
        } else {
            return TRUE;
        }

    } else {
        $this->errors['password'] = "Please provide a value for the password";
        return FALSE; 
    }

}

private function checkEmpty($string) {
    if (!empty($string)) {
        return TRUE;
    }

    return FALSE;
}

public function displayErrors() {
    $output = '';
    foreach ($this->errors as $error) {
        if (!empty($error)) {
            $output .= '<p>'.$error.'</p>';
        }
    }

    return $output;
}

 }

   ?>

   <?php

    require 'Validator.php';


      $validator = new Validator();


        $email = '';
        $password = '';


     if ($validator->checkPassword($password) && $validator->checkEmail($email)) {
echo 'You have entered a valid password and email.';
      } else {
echo $validator->displayErrors();
        }

      ?>

上面的代码来自两个单独的文件。从Validator.php开始的Validator来自Validator.php，而以require函数开头的那个来自index.php。所以我只是想知道为什么index.php中的$ validator-＆gt; displayErrors（）方法调用一次只显示一个错误，而不是一次显示所有错误。

Answer 1

替换

if ($validator->checkPassword($password) && $validator->checkEmail($email))

与

if ($validator->checkPassword($password) || $validator->checkEmail($email)) {

Answer 2

由于您的情况，只显示一个错误：

if ($validator->checkPassword($password) && $validator->checkEmail($email))

首先执行checkPassword方法，它返回false，因此永远不会检查第二个条件（应该执行第二个验证方法）。

您可以先通过执行验证方法来避免这种情况：

$validPassword = $validator->checkPassword($password);
$validEmail = $validator->checkEmail($email);

if ($validPassword && $validEmail) {
    echo 'You have entered a valid password and email.';
} else {
    echo $validator->displayErrors();
}

PHP类方法调用不显示所有错误

2 个答案: