define('anActionType', 1);
$actionTypes = array(anActionType => 'anActionType');
class core {
public $callbacks = array();
public $plugins = array();
public function __construct() {
$this->plugins[] = new admin();
$this->plugins[] = new client();
}
}
abstract class plugin {
public function registerCallback($callbackMethod, $onAction) {
if (!isset($this->callbacks[$onAction]))
$this->callbacks[$onAction] = array();
global $actionTypes;
echo "Calling $callbackMethod in $callbacksClass because we got {$actionTypes[$onAction]}" . PHP_EOL;
// How do I get $callbacksClass?
$this->callbacks[$onAction][] = $callbackMethod;
}
}
class admin extends plugin {
public function __construct() {
$this->registerCallback('onTiny', anActionType);
}
public function onTiny() { echo 'tinyAdmin'; }
}
class client extends plugin {
public function __construct() {
$this->registerCallback('onTiny', anActionType);
}
public function onTiny() { echo 'tinyClient'; }
}
$o = new core();
$callbacksClass
应该是管理员或客户。或者我完全忽略了这一点,应该采取另一种方式吗?应该注意的是,我只接受一个不要求我将classname作为参数发送到registerCallback方法的答案。
答案 0 :(得分:56)
如果有人来这里寻找如何从我这样的其他课程中获取通话类的名称,请查看https://gist.github.com/1122679
编辑:粘贴代码
function get_calling_class() {
//get the trace
$trace = debug_backtrace();
// Get the class that is asking for who awoke it
$class = $trace[1]['class'];
// +1 to i cos we have to account for calling this function
for ( $i=1; $i<count( $trace ); $i++ ) {
if ( isset( $trace[$i] ) ) // is it set?
if ( $class != $trace[$i]['class'] ) // is it a different class
return $trace[$i]['class'];
}
}
<强> EG 强>
class A {
function t() {
echo get_calling_class();
}
}
class B {
function x() {
$a = new A;
$a->t();
}
}
$b = new B;
$b->x(); // prints B
答案 1 :(得分:17)
使用get_class()
:
$this->callbacks[$onAction][] = $callbackMethod;
$className = get_class($this);
// Call callback method
$className->$callbackMethod();
答案 2 :(得分:4)
你应该做的事情如下:
$this->registerCallback(array($this, 'onTiny'), anActionType);
这就是PHP如何处理对象方法的句柄。