php - PHP : Trying to access array offset on value of type null

Question

这是错误来自的行...

If (password_verify($password, $user['password'])){}

这是声明中的内容...

    $_SESSION['id'] = $user['id']; 
    $_SESSION['username'] = $user['username']; 
    $_SESSION['email'] = $user['email']; 
    $_SESSION['verified'] = $user['verified']; 
    $_SESSION['message'] = "Success!"; 
    $_SESSION['alert-class'] = "alert-success"; 
    header ('location: verification.php'); 
    exit(); 
} else { 
    $errors['login_fail'] = "Incorrect Username or Password"; 
} 

Answer 1

您需要检查是否已设法从数据库加载用户。找不到用户时，从DB加载数据的函数看起来返回null。

您可以将条件更改为以下形式:

if(!empty($user) && password_verify($password, $user['password'])) {
  // ... log in user
} else {
  // ... do something when password or user doesn't match
}