我想构建一个更新配置文件页面,我有2个mysql表1.帐户2.信息,所以我想插入或更新用户信息表。
这是我到目前为止所做的,但在保存数据"Call to undefined method stdClass::save()"
控制器:
public function actionUpdate_profile() {
$model = new UserProfileForm();
$user = Userinfo::model()->findByPk(Yii::app()->user->id);
$model->name = $user->_name;
$model->myurl = $user->myurl;
if (isset($_POST['UserProfileForm'])) {
$model->attributes = $_POST['UserProfileForm'];
if ($model->validate()) {
$model->attributes = $_POST['UserProfileForm'];
$user->name = $model->name;
$user->myurl = $model->myurl;
$user->save();
});
});
$this->render('update_profile', array(
'model' => $model,
'user' => $user,
));
}
模型:
class Userinfo extends CActiveRecord {
public static function model($className = __CLASS__) {
return parent::model($className);
}
public function tableName() {
return 'user_info';
}
public function rules() {
return array(
array('email, name, myurl', 'length', 'max' => 255),
);
}
public function attributeLabels() {
return array(
'user_id' => Yii::t('yii', 'Id User'),
'name' => Yii::t('yii', 'First Name'),
'myurl' => Yii::t('yii', 'Last Name'),
);
}
class UserProfileForm extends CFormModel {
public $name;
public $myurl;
public function rules() {
return array(
array('name, myurl', 'required'),
);
}
public function attributeLabels() {
return array(
'name' => Yii::t('yii', 'First Name'),
'myurl' => Yii::t('yii', 'Last Name'),
);
}
}
答案 0 :(得分:0)
我知道这不是最佳做法,但它很简洁 -
if($user = Userinfo::model()->findByPk(Yii::app()->user->id)) {
//then work with $user here
}