i have a question :
when i open a popup window(validation information)
当我打开一个窗口的时候 弹出一个登录的窗口（验证信息，因为是未登录的状态）
from its parent window, after validation succeed, it should close
在输入验证信息后，关闭验证信息窗口，并且主窗口应该刷新成登录的状态
popup window,then refresh parent window( to login state),
but how should i do to refresh its parent window ( to login state)？
我在验证信息窗口 应该怎么做才能刷新主窗口为登录的状态呢？
(this is the code to open popup window :
/**
* 弹出的登录窗口（表单）
*/
public function actionPopLogin(){
// 登录(login)
$loginInfo = $this->getPost('LoginForm');
if($loginInfo){
$login = new LoginForm;
$login->attributes=$loginInfo;
if($login->validate() && $login->login()){
echo "<script>parent.ClosePop();</script>";//close popup window
//here,how to write code make parent window refresh to login state
//$this->redirect(Yii::app()->user->returnUrl);这样不行
}
}
$this->renderPartial('poplogin','',false,true);
}
any help be appreciated!
