PHP继承、静态方法和属性

No。这不是真的。静态方法和属性将与非静态方法和属性一样被继承，并遵循相同的可见性规则:

class A {
    static private $a = 1;
    static protected $b = 2;
    static public $c = 3;
    public static function getA()
    {
        return self::$a;
    }
}
class B extends A {
    public static function getB()
    {
        return self::$b;
    }
}
echo B::getA(); // 1 - called inherited method getA from class A
echo B::getB(); // 2 - accessed inherited property $b from class A
echo A::$c++;   // 3 - incremented public property C in class A
echo B::$c++;   // 4 - because it was incremented previously in A
echo A::$c;     // 5 - because it was incremented previously in B

后两个是显著差异。在基类中增加继承的静态属性也会在所有子类中增加该属性，反之亦然。

没有(显然我在问题中看不到不是)。public和protected的静态方法和属性按照您期望的方式继承:

<?php
class StackExchange {
    public static $URL;
    protected static $code;
    private static $revenue;
    public static function exchange() {}
    protected static function stack() {}
    private static function overflow() {}
}
class StackOverflow extends StackExchange {
    public static function debug() {
        //Inherited static methods...
        self::exchange(); //Also works
        self::stack();    //Works
        self::overflow(); //But this won't
        //Inherited static properties
        echo self::$URL; //Works
        echo self::$code; //Works
        echo self::$revenue; //Fails
    }
}
StackOverflow::debug();
?>

静态属性和方法遵循可见性和继承规则，如下段所示。