Question

php高手来帮忙解答下，define('ROUTE_M', $this->param->route_m());

class application {
private $param;
public function __construct() {
global $_cfg;
$this->param =&System::load_sys_class('param');//获取路径 ，疑问1：是否执行了&System::load_sys_class('param')该语句然后将路径赋值给param？
define('ROUTE_M', $this->param->route_m()); //疑问2：怎么使用了两个->怎么理解？
define('ROUTE_C', $this->param->route_c());
define('ROUTE_A', $this->param->route_a());
$_cfg['route_m'] = ROUTE_M;
$_cfg['route_c'] = ROUTE_C;
$_cfg['route_a'] = ROUTE_A;
$this->global_start();
$this->global_init();
$this->global_end();

}

public function route_m() {

if(empty($this->route_url[1])){
$this->route_url[1]=$this->route_config['default']['m'];
}
define('G_MODULE_PATH',WEB_PATH.'/'.$this->route_url[1]);
return $this->route_url[1];
}
}

Answer 1

::表示执行静态方法。
$this->param->route_m()表不执行$this->param里的route_m()
在一个类里，$this->param表是类中的本类中的$param
假如route_m()是类aaa里面方法
$this->param=new aaa();
然后就可以执行
$this->param->route_m()