Question

我正在尝试使用Doctrine2将我的PHP应用程序移植到Zend Framework。这是一个网上商店平台。我需要编写一个带有额外条件的DQL查询来从用户购物车中的产品中选择附件。

我有一个传统的查询：

SELECT 
    ... 
FROM 
    product 
INNER JOIN 
    accessory_to_product ON(product.productid = accessory_to_product.accessoryid AND accessory_to_product.productid IN(000,000,000))

但这必须是主义。我有以下实体：

<?php

/**
 * @ORM\Table(name="product")
 * @ORM\Entity(repositoryClass="BestBuy\Entity\Repository\ProductRepository")
 */
class Product extends \BasicEntity
{

    /**
     * 
     * @var integer
     * @ORM\Column(type="integer",nullable=false)
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="IDENTITY")
     */
    protected $productid;

    /**
     *
     * @var string
     * @ORM\Column(type="string",nullable=false)     
     */
    protected $title;

    /**
     *
     * @var string
     * @ORM\Column(type="text",nullable=false)   
     */
    protected $description;

    /**
     *
     * @var string
     * @ORM\Column(type="datetime",nullable=false)   
     */
    protected $insertdate;

    //....

    /**
     *
     * @var array
     * 
     * @ORM\ManyToMany(targetEntity="Product")
     * @ORM\JoinTable(name="accessory_to_product",
     *  joinColumns={@ORM\JoinColumn(name="accessoryid", referencedColumnName="productid")},
     *  inverseJoinColumns={@ORM\JoinColumn(name="accessoryid", referencedColumnName="productid", unique=false)});
     */    
    protected $accessories;




}

因此我想将此查询重写为Doctrine2：

$qb = $this->getEntityManager()->createQueryBuilder();
$qb->select('p')
    ->from('BestBuy\Entity\Product', 'p')
    ->join('p.accessories', 'a', Expr\Join::WITH, $qb->expr()->in('p.productid', ':products'))
    ->setParameter('products', $products);

但是这给了我以下查询：

SELECT 
    p0_.productid AS productid0, 
    p0_.title AS title1, 
    p0_.description AS description2, 
    p0_.insertdate AS insertdate3, 
    p0_.price AS price4, 
    p0_.model AS model5, 
    p0_.categoryid AS categoryid6, 
    p0_.deliverytimeid AS deliverytimeid7, 
    p0_.status AS status8, 
    p0_.weight AS weight9, 
    p0_.pagetitle AS pagetitle10, 
    p0_.metadescription AS metadescription11, 
    p0_.youtubecode AS youtubecode12 
FROM 
    product p0_ 
INNER JOIN 
    accessory_to_product a2_ ON p0_.productid = a2_.accessoryid 
INNER JOIN 
    product p1_ ON p1_.productid = a2_.accessoryid AND (p0_.productid IN (11224))

除了将accessory_to_product表连接两次外，这几乎是正确的。有没有人给我一个提示？实际上我应该选择FROM accessoires然后加入产品，但由于这是与产品的关系，这是不可能的。

Answer 1

终于搞定了。我的关系不正确：

/**
*
* @var array
* 
* @ORM\ManyToMany(targetEntity="Product", fetch="EXTRA_LAZY")
* @ORM\JoinTable(name="accessory_to_product",
*  joinColumns={@ORM\JoinColumn(name="accessoryid", referencedColumnName="productid")},
*  inverseJoinColumns={@ORM\JoinColumn(name="accessoryid", referencedColumnName="productid", unique=false)});
*/

需要：

/**
*
* @var array
* 
* @ORM\ManyToMany(targetEntity="Product", fetch="EXTRA_LAZY")
* @ORM\JoinTable(name="accessory_to_product",
*  joinColumns={@ORM\JoinColumn(name="accessoryid", referencedColumnName="productid")},
*  inverseJoinColumns={@ORM\JoinColumn(name="productid", referencedColumnName="productid", unique=false)});
*/

因为accessory_to_product表有两列，accesoryid和productid。以下代码：

    $qb = $this->getEntityManager()->createQueryBuilder();
    $qb->select('p')
            ->from('BestBuy\Entity\Product', 'p')
            ->where($qb->expr()->in('a.productid', ':products'))    
            ->leftJoin('p.accessories', 'a')
            ->setParameter('products', $products);

在SQL中进行以下查询：

SELECT 
    p0_.productid AS productid0, 
    p0_.title AS title1, 
    p0_.description AS description2, 
    p0_.insertdate AS insertdate3, 
    p0_.price AS price4,
    p0_.model AS model5, 
    p0_.categoryid AS categoryid6, 
    p0_.deliverytimeid AS deliverytimeid7, 
    p0_.status AS status8, 
    p0_.weight AS weight9, 
    p0_.pagetitle AS pagetitle10, 
    p0_.metadescription AS metadescription11, 
    p0_.youtubecode AS youtubecode12 
FROM 
    product p0_ 
LEFT JOIN 
    accessory_to_product a2_ ON p0_.productid = a2_.accessoryid 
LEFT JOIN 
    product p1_ ON p1_.productid = a2_.productid 
WHERE 
    p1_.productid IN (11224)

与原版略有不同，但会给出完全相同的结果！

加入自我引用manyToMany关系的条件

1 个答案: