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我想将对象包装在另一个对象中 - 偏爱合成而不是继承。但我不确定我是否做对了没有错误。问题abt PHP魔术方法和授权
我创建了一个类Wrapped
这就是包装Wrapper
。我在$wrapper
上调用方法/属性时,如果它存在于类Wrapper
中,它将返回else,它将委托给$wrapped
对象。我想知道除了事实我没有检查方法/财产是否存在,我做错了什么?可以解释一下__callStatic()
吗?
class Wrapped {
protected $wrappedProp1 = 'Wrapped: Property 1';
protected $wrappedProp2 = 'Wrapped: Property 2';
function method1($arg1, $arg2) {
echo "Called Wrapped::method1() with the following parameters: $arg1, $arg2";
}
static function sMethod2() {
echo 'Called a static method in wrapped';
}
function __get($name) {
return $this->$name;
}
function __set($name, $val) {
$this->$name = $val;
}
}
class Wrapper {
protected $wrapperProp1 = 'Wrapper: Property 1';
protected $wrapped;
function __construct($wrapped) {
$this->wrapped = $wrapped;
}
function wrapperMethod() {
echo 'In wrapper method';
}
function __get($name) {
if (property_exists($this, $name)) {
return $this->$name;
}
return $this->wrapped->$name;
}
function __set($name, $val) {
if (property_exists($this, $name)) {
$this->$name = $val;
}
$this->wrapped->$name = $val;
}
function __call($name, $args = array()) {
call_user_func_array(array($this->wrapped, $name), $args);
}
static function __callStatic($name, $args = array()) {
call_user_func_array(array('Wrapped', $name), $args);
}
}
$wrapper = new Wrapper(new Wrapped);
// testing normal methods
$wrapper->wrapperMethod();
echo $wrapper->wrapperProp1;
$wrapper->wrapperProp1 = 'New Wrapper Prop 1';
echo $wrapper->wrapperProp1;
// testing delegates
$wrapper->method1('hello', 'world'); //delegated to Wrapped::method1()
$wrapper->sMethod2(); // delegated to static Wrapped::sMethod2() ... what is callStatic for then
echo $wrapper->wrappedProp2;
Wrapper::sMethod2();