[ACCEPTED]-findByExample in Doctrine-doctrine

Accepted answer
Score: 25

You can use the findBy method, which is inherited 11 and is present in all repositories.

Example:

$criteria = array('name' => 'someValue', 'status' => 'enabled');
$result = $em->getRepository('SomeEntity')->findBy($criteria);

You 10 can create findByExample method in one of your repositories 9 using a definition like this:

class MyRepository extends Doctrine\ORM\EntityRepository {
    public function findByExample(MyEntity $entity) {
        return $this->findBy($entity->toArray());
    }
}

In order for 8 this to work, you will have to create your 7 own base class for the entities, implementing 6 the toArray method.

MyEntity can also be an interface, which 5 your specific entities will have to implement 4 the toArray method again.

To make this available 3 in all your repositories, ensure that you 2 are extending your base repository class 1 - in this example, the MyRepository one.

P.S I assume you are talking about Doctrine 2.x

Score: 10

Yes.

Let's say you have a model called 2 Users. You have the following two classes

abstract class Base_User extends Doctrine_Record 
{
   //define table, columns, etc
}

class User extends Base_User
{

}

in 1 some other object you can do

$user = new User;

//This will return a Doctrine Collection of all users with first name = Travis
$user->getTable()->findByFirstName("Travis");

//The above code is actually an alias for this function call
$user->getTable()->findBy("first_name", "Travis");

//This will return a Doctrine Record for the user with id = 24
$user->getTable()->find(24);

//This will return a Doctrine Collection for all users with name=Raphael and 
//type = developer
$user->getTable()
     ->findByDql("User.name= ? AND User.type = ?", array("Raphael", "developer"));
Score: 3
$users = $userTable->findByIsAdminAndIsModeratorOrIsSuperAdmin(true, true, true);

See http://www.doctrine-project.org/projects/orm/1.2/docs/manual/dql-doctrine-query-language/en

0

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