打开PHP5对象，不需要先声明变量名

Posted 2021-02-27

I use this object verry often, so that I only need to pass one object as parameter on any function. No need to wory when we refactore the function
Usage:
$var = new open();
$var->url = 'http://www.world.com';
$var->title = 'Testing Site';

echo some_function($var);

//php5 only//
 
class open
{
   var $____x = array();
   function __construct($a = array())
   {
      $this->fromArray($a);
   }
 
   function __set($name,$val)
   {
      $this->____x[$name] = $val;
   }
 
   function __get($name)
   {
       if(isset($this->____x[$name])) return $this->____x[$name];
   }
 
   function __isset($name)
   {
      return isset($this->____x[$name]);
   }
 
   function toArray()
   {
      return $this->____x;
   }
 
   function fromArray($a = array())
   {
      if(count($a)>0) foreach($a as $k => $v) $this->____x[$k] = $v;
   }
 
}