findall() in yii return array and i want it to return object how - yii

Question

i have w function called update that i want to update a specific record in database according to module id (m_id) and key (key)
here is the function
public function updateRecord($module,$key,$newData){
$id = $this->module($module);
$model = new Form;
$criteria = new CDbCriteria();
$criteria->condition = "`m_id` = 1 AND `key` = 'txt'";
// $model = Yii::app()->db->createCommand("SELECT * FROM `tbl_setting` WHERE `m_id` = 1 AND `key` = 'txt'")->query();
die(var_dump($model->findAll($criteria)));
// $model = Yii::app()->db->createCommand("SELECT * FROM `tbl_setting` WHERE `m_id` = 1 AND `key` = 'txt'")->execute();
$model->m_id = $id ;
$model->attributes = $newData;
$model->save();
//die(print_r($model['m_id']));
//if(isset($_POST['save'])){
//$model['m_id'] = $id;
//}
}
every time i try to use this function it give me error that $model is not an object so it can't execute
$model->m_id = $id ;
i check out the type of findAll() and it gives an array
i checkout all find() methods and only findByPk() that give an object
and i want to choose according to the $criteria and i don't know what to do any help ?! :)

Answer 1

findAll will always return an array. With objects if there are record found, or an empty array on no result. find only returns 1 record, or NULL if nothing is found:
http://www.yiiframework.com/doc/api/1.1/CActiveRecord#find-detail
Your code will be save when you do it like this:
if(!is_null($model)) {
$model->m_id = $id ;
$model->attributes = $newData;
$model->save();
} else {
echo 'No record found';
}