php 教义中的 findByExample
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findByExample in Doctrine
提问by rizidoro
Is there a method in Doctrinelike Hibernate's findByExamplemethod?
Doctrine 中是否有类似Hibernate的findByExample方法的方法？
thanks
谢谢
回答by Nikola Petkanski
You can use the findBymethod, which is inherited and is present in all repositories.
您可以使用该findBy方法，该方法是继承的并且存在于所有存储库中。
Example:
例子：
$criteria = array('name' => 'someValue', 'status' => 'enabled');
$result = $em->getRepository('SomeEntity')->findBy($criteria);
You can create findByExamplemethod in one of your repositories using a definition like this:
您可以findByExample使用如下定义在您的存储库之一中创建方法：
class MyRepository extends Doctrine\ORM\EntityRepository {
public function findByExample(MyEntity $entity) {
return $this->findBy($entity->toArray());
}
}
In order for this to work, you will have to create your own base class for the entities, implementing the toArraymethod.
为了使其工作，您必须为实体创建自己的基类，实现该toArray方法。
MyEntitycan also be an interface, which your specific entities will have to implement the toArraymethod again.
MyEntity也可以是一个接口，您的特定实体必须toArray再次实现该方法。
To make this available in all your repositories, ensure that you are extending your base repository class - in this example, the MyRepositoryone.
要使其在您的所有存储库中可用，请确保您正在扩展您的基本存储库类 - 在本例中，是MyRepository一个。
P.S I assume you are talking about Doctrine 2.x
PS 我假设你在谈论Doctrine 2.x
回答by Travis
Yes.
是的。
Let's say you have a model called Users. You have the following two classes
假设您有一个名为 Users 的模型。你有以下两个类
abstract class Base_User extends Doctrine_Record
{
//define table, columns, etc
}
class User extends Base_User
{
}
in some other object you can do
在其他一些你可以做的对象中
$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"));
回答by Carlo Lopez
$users = $userTable->findByIsAdminAndIsModeratorOrIsSuperAdmin(true, true, true);
See http://www.doctrine-project.org/projects/orm/1.2/docs/manual/dql-doctrine-query-language/en
见http://www.doctrine-project.org/projects/orm/1.2/docs/manual/dql-doctrine-query-language/en
