Access Level to certain class must be public error in PHP

Question

I created this class

Answer 1

Youre getting that error because the visibility of the method must be the same or less restrictive than that of it its definition on a parent class. In this case you have addErrors as public on your abstract class and are attempting to make it protected on a child class.

Answer 2

Since PHP 7.2.0 there was a bug #61970 fixed (Restraining __construct() access level in subclass gives a fatal error).

Bump your PHP version to 7.2 and you could make it private or protected.

Answer 3

You've specify the protected access to the protected function _addErrors(array $errors) method of Form_Element_Validators class. So change it to public.

Edit:

Have you noticed? The sub class method (overridden method) is defined with Type Hinting. Please keep the same parameter type for both; super-class and sub-class method.

 abstract class Validator{
        public $_errors = array();
        abstract public function isValid($input);

        public function _addErrors(array $message){
            $this->_errors = $message;
        }
        ....

Answer 4

It also happens when you are trying to do something with specific guard. You can add

 protected $guard = "guard_name";

Add above line in your model. that will solve the problem.

Answer 5

As others have mentioned, you can't make a sub class method more restrictive than the parent; this is because sub classes are supposed to be a valid replacement for their parent class.

In your particular case, I would change the visibility of all methods and properties that start with an underscore to protected.