In my application, i have fields that are common to all tables, like create date, update date etc. To assign these values i'm using beforeValidate callback. Now, this callback is same for all models.
To avoid code duplication, i want to create a base model class.
But, when I tried to create a base model, yii thrown error saying table cannot be found in database, which is true since I dont have any table for this base model.
Is there any way I can create a base model class.
Yes, if you work with dynamic DB structure or have other reasons to work with Yii ActiveRecord without creating classes for each table in DB, you may use smartActiveRecord yii extension
I separated it few minuts ago from my other extension -- AR behavior that adds versioning to any model (it copies all data on insert & update to special table (and create it if it's absent), that have a same structure as source table + "revision field" and primary key extended by this field.
Look at SmartAR.php source, there is example of usage in comments.
Take a look at CTimeStampBehavior.
Incase that doesn't help you, you can just write a behavior class yourself.
Hope this helps.
Edit:
Assuming you are using ActiveRecords.
If you want to create a new base model, you can do this:
abstract class MyBaseARClass extends CActiveRecord{
protected function beforeValidate(){
if(parent::beforeValidate()){
// assign your fields
return true;
}
else return false;
}
}
Have you created a base table? Thinking about the Yii framework it may be easier to have a relationship between a model and the base model.
In your case, you need to override
public static function model($className=__CLASS__)
{
return parent::model($className);
}
in every child class so Yii would know which table to use for your model. Otherwise it will try and use base class as table name.
I.e.
class User extends BaseActiveRecord {
public static function model($className=__CLASS__)
{
return parent::model($className);
}
}
