converting this query to yii CDbCriteria - mysql

Question

I'm pretty new to yii, and still trying to figure out some things. How do i change this mysql query to Yii's query using CDbCriteria?
SELECT DISTINCT p.prefix, p.state
FROM `store` as d
JOIN (`zip` as p)
ON (d.zip = p.zip)
WHERE d.store_code='".(int)$storeCode."'
GROUP BY p.prefix, p.state
ORDER BY p.state ASC
This is what i've done so far
$criteria = new CDbCriteria();
$criteria->select = array('prefix','state');
$criteria->join="zip";
$criteria->condition = 'store_code=:store_code';
$criteria->params = array(':store_code'=> (int)$storeCode);
$criteria->order = 'state ASC';
$query = AB::model()->findAll($criteria); //query
but it gives me this error
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'zip WHERE store_code=64 ORDER BY state ASC' at line 1. The SQL statement executed was: SELECT prefix, state FROM `store` `t` zip WHERE store_code=:store_code ORDER BY state ASC. Bound with :store_code=64
BONUS: when i use the 1st query with the command below. it takes 2.2 seconds to return a result. any idea why its so slow? or how to speed it up? (the store table has 2518 rows and state has 20 rows)
$query = Yii::app()->db->createCommand($sql)->queryAll();

Answer 1

To answer your immediate question, I think the join syntax is off. From the docs:
how to join with other tables. This refers to the JOIN clause in an SQL statement. For example, 'LEFT JOIN users ON users.id=authorID'.
So in our case your command should run by doing something like
$criteria = new CDbCriteria();
$criteria->select = array('prefix','state');
$criteria->join="LEFT JOIN zip ON t.zip = zip.zip";
$criteria->condition = 'store_code=:store_code';
$criteria->params = array(':store_code'=> (int)$storeCode);
$criteria->order = 'state ASC';
$query = AB::model()->findAll($criteria); //query
t is what Yii aliases your base table with (whatever AB is representing. store?).
You can also use the Yii query builder to accomplish this task. For example
$data = Yii::app()->db->createCommand()
->selectDistinct(array("prefix","store"))
->from('store d')
->leftJoin('zip p', 'd.zip=p.zip')
->where('store_code=:store_code', array(':store_code'=>(int)$storeCode))
->queryAll();
BUT, I'd highly recommend looking more into relationships of models. Yii's active record is pretty sweet. If you build relationships, all you need to do is find the models you want with your AB::models->findAllByAttributes(array("store_code"=>$myCode));, and do $abModel->zip->column_name
You can also do something called 'eager loading'. Depending on how many times you plan on querying the relationship, (in a for loop, or just once) you may want to eager load to boost performance. E.g AB::model()->with("zip")->findAll($criteria);
You don't have to, and it's not always best practice to eager load.
If you build relationships, all you need to do is something like
$abModels = AB::models->with('zip')->findAllByAttributes(array("store_code"=>$myCode));
foreach ($abModels as $abModel) {
$state = $abModel->zip->state;
$prefix = $abModel->zip->prefix;
}
Active Record relationships
Active Record
Query Builder
Cheers

Answer 2

Try this:
$criteria = new CDbCriteria();
$criteria->alias = 'd';
$criteria->select = array('p.prefix','p.state');
$criteria->join='JOIN zip d ON d.zip = p.zip';
$criteria->condition = 'd.store_code=:store_code';
$criteria->params = array(':store_code'=> (int)$storeCode);
$criteria->group='p.prefix, p.state';
$criteria->order = 'p.state ASC';