Doctrine 2 ArrayCollection filter method

荒凉一梦 提交于 2019-11-26 17:59:12

问题


Can I filter out results from an arrayCollection in Doctrine 2 while using lazy loading? For example,

// users = ArrayCollection with User entities containing an "active" property
$customer->users->filter('active' => TRUE)->first()

It's unclear for me how the filter method is actually used.


回答1:


The Boris Guéry answer's at this post, may help you: Doctrine 2, query inside entities

$idsToFilter = array(1,2,3,4);

$member->getComments()->filter(
    function($entry) use ($idsToFilter) {
       return in_array($entry->getId(), $idsToFilter);
    }
); 



回答2:


Doctrine now has Criteria which offers a single API for filtering collections with SQL and in PHP, depending on the context.

https://www.doctrine-project.org/projects/doctrine-orm/en/latest/reference/working-with-associations.html#filtering-collections

Update

This will achieve the result in the accepted answer, without getting everything from the database.

use Doctrine\Common\Collections\Criteria;

/**
 * @ORM\Entity
 */
class Member {
  // ...
  public function getCommentsFiltered($ids) {
    $criteria = Criteria::create()->where(Criteria::expr()->in("id", $ids));

    return $this->getComments()->matching($criteria); 
  }
}



回答3:


Your use case would be :

    $ArrayCollectionOfActiveUsers = $customer->users->filter(function($user) {
                        return $user->getActive() === TRUE;
                    });

if you add ->first() you'll get only the first entry returned, which is not what you want.

@ Sjwdavies You need to put () around the variable you pass to USE. You can also shorten as in_array return's a boolean already:

    $member->getComments()->filter( function($entry) use ($idsToFilter) {
        return in_array($entry->getId(), $idsToFilter);
    });



回答4:


The Collection#filter method really does eager load all members. Filtering at the SQL level will be added in doctrine 2.3.



来源:https://stackoverflow.com/questions/8334356/doctrine-2-arraycollection-filter-method

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!