我如何在yii中使用ajax来显示针对id的结果?
public function actionView(){
$model= new ViewForm();
$model->unsetAttributes();
if (isset($_GET['ViewJob'])) {
$model->attributes = $_GET['ViewJob'];
}
$this->render('viewjob',array(
'model'=>$model
));
}
答案 0 :(得分:0)