静态在php中，为什么它为这两个类使用一个缓存

Question

I create parent class for my models, it look like this

class Model {

   protected static $cache = [];

   public static function load($id)
   {
       return static::$cache[$id];
   }

   public static function findById($id)
   {
       $model = static::find($id);

       static::$cache[$id] = model;

       return model;
   }

   ...
}

And i have two child class, examples

class A extend Model {}
class B extend Model {}

then, i write

A::findById(1);
B::findById(1);

$a = A::load(1);
$b = B::load(1);

I think type of class $b is class B, but not, it is class A, because function "load" use static cache and load result from function A::findById(1). Why?

Answer 1

By accessing static class properties as static::$cache you tell the interpreter to use the $cache property of the class used to call the method and not the one of the class where the method is defined.

In plain English, when you call A::findById() you expect that findById() uses the property $cache of A and not the one of Model. The same for A::load() and similar for B::findById() and B::load().

The question now is: do the A and B classes declare a static property $cache on their own? Because if they do not declare this property, the static:: in front of $cache in the function A::findById() cannot find A::$cache and uses Model::$cache instead. The same for B::findById().

The easy solution to your problem is to declare the protected static $cache property in all classes that extend Model.

The correct solution is to not use static properties and methods. Static properties and methods are not OOP. They are procedural programming (and thinking) disguised as OOP code. Code that is difficult to understand and test.