如何自我初始化Doctrine_Record（就好像Doctrine_Query那样）？

Question

I have models that extend Doctrine_Record, and are directly mapped to one specific record from the database (that is, one record with given a given id, hardcoded statically in the class).

Now, I want the specific record class to initialize itself as if Doctrine_Query would. So, this would be the normal procedure:

$query = new Doctrine_Query();
$model = $query->from('Model o')->where('id = ?', 123)->fetchOne();

I would like to do something like this

$model = new Model();

And in the Model:

const ID = 123;

//note that __construct() is used by Doctrine_Record so we need construct() without the __
public function construct()
{
    $this->id = self::ID;
    //what here??
    $this->initialize('?????');
}

So for clarity's sake: I would like the object to be exactly the same as if it would be received from a query (same state, same attributes and relations and such).

Any help would be greatly appreciated: thank you.

Answer 1

The first thing I need to say is I'd put the constant in the class. So like this:

class Application_Model_Person
{
    const ID = 1234;
}

Then, a Doctrine method like Doctrine_Record::fetchOne() is always returning a (new) instance of the model and never merges the data with the record you're calling fetchOne() to. Doctrine is nevertheless able to merge a retreived record with another class, so it rather simple to do:

class Application_Model_Person extends Doctrine_Record_Abstract
{
    const ID = 1234;

    public function __construct($table = null, $isNewEntry = false)
    {
        // Calling Doctrine_Record::__construct
        parent::__construct($table, $isNewEntry);

        // Fetch the record from database with the id self::ID
        $record = $this->getTable()->fetchOne(self::ID);
        $this->merge($record);
    }
}

Then you're able to do:

$model = new Application_Model_Person;
echo $model->id; // 1234