我有三个实体 Travel、Favorite 和 User,每个用户都可以将旅行添加到他的收藏列表中。
当显示所有旅行列表时,如果旅行已经在用户的收藏夹中,则显示一个“移除”按钮;如果不在列表中,则显示“添加到收藏夹”按钮。
在操作控制器中,我使用查询构建器获取当前已认证用户的所有收藏列表,并将结果放入 ($favorite) 变量中。
public function listAction($page, Request $request)
{
$em = $this->getDoctrine()->getManager();
// code of boutton add to or remove from favorite
$user = $this->getUser(); // this is a function to verify if user is AUTHENTICATED_REMEMBERED
if($user) {
$favorite = $em->getRepository('ProjectTravelBundle:Favorite')->getFavoriteByUser($user);
}
else{
$favorite = '';
}
$paginator = $this->get('knp_paginator');
$qb = $em->getRepository('ProjectTravelBundle:Travel')->getListTravelsFrontend();
$nb = $qb->getQuery()->getResult();
$pagination = $paginator->paginate(
$qb,
$request->query->get('page', $page),10);
return $this->render('ProjectFrontendBundle:Travel:travel-list-view.html.twig',array(
'pagination' => $pagination,
'nb' => $nb,
'favorite' =>$favorite, // list of favorites of current user
));
}
我认为的解决方案是测试旅行是否在数组收藏夹中 (favorite),如果是则显示删除按钮,否则显示添加按钮,但我无法在 show.twig 中实现这一点。
{% for travel in pagination %}
//..................
{% if is_granted('IS_AUTHENTICATED_REMEMBERED') %}
{% if travel.id not in favorite %}
<span><a href="{{ path('frontend_travel_add_favorite', {'id': travel.id} ) }}" class="pull-right button btn-small red">Add to favorite</a></span>
{% else %}
<span><a href="{{ path('frontend_travel_delete_favorite', {'id': favorite.id} ) }}" class="pull-right button btn-small red">Remove from</a></span>
{% endif %}
// if user is not AUTHENTICATED_REMEMBERED
{% else %}
<span><a href="{{ path('frontend_travel_add_favorite', {'id': travel.id} ) }}" class="pull-right button btn-small red">Add to favorite</a></span>
{% endif %}
//..........
{% endfor %}
代码库: queryBuilder
// get list of favorite of current user
public function getFavoriteByUser($user)
{
$qb = $this->createQueryBuilder('f')
->Where('f.user = '.$user);
return $qb->getQuery()->getResult();
}
更新
FavoriteEntity(收藏实体)
class Favorite
{
/**
* @var integer
*
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @ORM\ManyToOne(targetEntity="Project\UserBundle\Entity\User")
* @ORM\JoinColumn(nullable=false)
*/
protected $user;
/**
* @ORM\ManyToOne(targetEntity="Project\TravelBundle\Entity\Travel", inversedBy="favorites")
* @ORM\JoinColumn(nullable=false)
*/
protected $travel;
/**
* Get id
*
* @return integer
*/
public function getId()
{
return $this->id;
}
/**
* Set user
*
* @param \Project\UserBundle\Entity\User $user
* @return Favorite
*/
public function setUser(\Project\UserBundle\Entity\User $user)
{
$this->user = $user;
return $this;
}
/**
* Get user
*
* @return \Project\UserBundle\Entity\User
*/
public function getUser()
{
return $this->user;
}
/**
* Set travel
*
* @param \Project\TravelBundle\Entity\Travel $travel
* @return Favorite
*/
public function setTravel(\Project\TravelBundle\Entity\Travel $travel)
{
$this->travel = $travel;
return $this;
}
/**
* Get travel
*
* @return \Project\TravelBundle\Entity\Travel
*/
public function getTravel()
{
return $this->travel;
}
}
TravelEntity
class Travel
{
/....
/**
* @ORM\OneToMany(targetEntity="Project\TravelBundle\Entity\Favorite", mappedBy="travel", cascade={"remove"})
*/
private $favorites;
public function __construct()
{
$this->favorites = new \Doctrine\Common\Collections\ArrayCollection();
}
/...........
/**
* Add favorites
*
* @param \Project\TravelBundle\Entity\Favorite $favorites
* @return Travel
*/
public function addFavorite(\Project\TravelBundle\Entity\Favorite $favorites)
{
$this->favorites[] = $favorites;
$favorites->setTravel($this);
return $this;
}
/**
* Remove favorites
*
* @param \Project\TravelBundle\Entity\Favorite $favorites
*/
public function removeFavorite(\Project\TravelBundle\Entity\Favorite $favorites)
{
$this->favorites->removeElement($favorites);
}
/**
* Get favorites
*
* @return \Doctrine\Common\Collections\Collection
*/
public function getFavorites()
{
return $this->favorites;
}
$favorites
的OneToMany关系。这样的关系可能会消耗大量资源。 - Ziumin