爲什麼symfony爲連接執行額外的查詢

Question

我嘗試用myisam表連接many_many從一個數據庫加入myisam表。

我有表文章，article_category，演員，article_actor（和許多更類似many2many關係）。

我在

條安裝：

class Article{ 
//.... 

/** 
* @ORM\ManyToMany(targetEntity="Category", inversedBy="articles") 
* @ORM\JoinTable(name="article_category") 
*/ 
protected $categories; 

/** 
* @ORM\ManyToMany(targetEntity="Actor", inversedBy="articles") 
* @ORM\JoinTable(name="article_actor") 
*/ 
protected $actors; 

/** 
* @ORM\ManyToMany(targetEntity="Cameraman", inversedBy="cameramen") 
* @ORM\JoinTable(name="article_cameraman") 
*/ 
public function __construct() { 
    $this->categories = new ArrayCollection(); 
    $this->actors = new ArrayCollection(); 
} 

類別：

類分類{

/** 
* @ORM\ManyToMany(targetEntity="Article", mappedBy="categories") 
**/ 
protected $articles;  

public function __construct() { 
    $this->articles = new ArrayCollection(); 
} 

演員：

類演員{

/** 
* @ORM\ManyToMany(targetEntity="Article", mappedBy="actors") 
**/ 
protected $articles; 

public function __construct() { 
    $this->articles = new ArrayCollection(); 
} 

在ArticleRepository我甲肝查詢與聯接：

類ArticleRepository延伸EntityRepository {
個公共職能findArticleWithJoins（$ _ titleSlug）{

$q = $this->createQueryBuilder('a') 
    ->where('a.titleSlug = :titleSlug') 
    ->setParameter('titleSlug', $_titleSlug) 
    ->leftJoin('a.categories', 'c') 
    ->leftJoin('a.actors', 'ac') 


return $q->getQuery()->getSingleResult(); 

} }

在我的控制器我取與相關的文章加入：

$em = $this->getDoctrine()->getManager(); 
$article = $em->getRepository('MyBundle:Article')->findArticleWithJoins('theSlug'); 
return $this->render('MyBundle:Default:index.html.twig',array('article' => $article)); 

，最後在模板一旦我遍歷連接，有3個查詢：

{% block body %} 
    {{article.title}}<br/> 
    {{article.categories.0.name}}<br/> 
    {% for actor in article.actors %} 
    <li>{{ actor.firstName }} {{ actor.lastName }}</li> 
    {% endfor %} 
{% endblock %} 

第QUER是：

SELECT 
    a0_.id AS id0, 
    a0_.title AS title15, 
    a0_.orig_title AS orig_title16, 
    a0_.title_slug AS title_slug17, 

FROM 
    article a0_ 
    LEFT JOIN article_category a2_ ON a0_.id = a2_.article_id 
    LEFT JOIN category c1_ ON c1_.id = a2_.category_id 
    LEFT JOIN article_actor a4_ ON a0_.id = a4_.article_id 
    LEFT JOIN actor a3_ ON a3_.id = a4_.actor_id 
WHERE 
    a0_.title_slug = ? 

第二個查詢：

SELECT 
    t0.id AS id1, 
    t0.name AS name2, 
    t0.slug AS slug3, 
FROM 
    category t0 
    INNER JOIN article_category ON t0.id = article_category.category_id 
WHERE 
    article_category.article_id = ? 

而第三個查詢：

SELECT 
    t0.id AS id1, 
    t0.first_name AS first_name2, 
    t0.last_name AS last_name3, 
    t0.name_slug AS name_slug4, 
FROM 
    actor t0 
    INNER JOIN article_actor ON t0.id = article_actor.actor_id 
WHERE 
    article_actor.article_id = ? 

我想避免多個查詢通過使用該連接findArticleWithJoins（）函數。 但結果是相同的只是查詢文章，然後選擇關係。

我做錯了什麼？

Answer 1

添加選擇（->select('a, c, ac')到您的查詢：。

$q = $this->createQueryBuilder('a') 
    ->select('a, c, ac') 
    ->where('a.titleSlug = :titleSlug') 
    ->setParameter('titleSlug', $_titleSlug) 
    ->leftJoin('a.categories', 'c') 
    ->leftJoin('a.actors', 'ac') 

這將使學說建構的對象與關係