我正在使用querybuilder应用以下查询，但不知何故，很多2关系没有按预期工作。
$shopData = $sm->createQueryBuilder()
->select('v')
->from('AdminBundle:Voucher','v')
->innerJoin('v.shop', 's')
->leftJoin('AdminBundle:VoucherProgram', 'vp', \Doctrine\ORM\Query\Expr\Join::ON, 'vp.id = v.program_id')
->leftJoin('AdminBundle:shopHistory', 'sh', \Doctrine\ORM\Query\Expr\Join::ON, 'sh.shop = s.id')
->where('s.shopStatus = :shopStatus')
->setParameter('shopStatus', Shop::SHOP_ACTIVATED)
->andWhere('s.highlightedHome = :highlightedHome')
->setParameter('highlightedHome', Shop::SHOP_HIGHLIGHTED_HOME)
->andWhere('s.offers = \'voucher\'')
->setFirstResult(0)
->setMaxResults(6)
->addOrderBy('v.discount_amount', 'DESC')
->groupBy('sh.shop')
->getQuery()
->getSql();
生成的查询如下所示：
SELECT v FROM AdminBundle:Voucher v INNER JOIN v.shop s LEFT JOIN AdminBundle:VoucherPrograms vp ON vp.id = v.program_id LEFT JOIN AdminBundle:shopHistory sh ON sh.shop = s.id WHERE s.shopStatus = :shopStatus AND s.highlightedHome = :highlightedHome AND s.offers = 'voucher' GROUP BY sh.shop ORDER BY v.discount_amount DESC
如果我删除了所有内容并且只保留了Many2Many关系的内连接，那么它正在按预期工作。
$sm->createQueryBuilder()
->select('v')
->from('AdminBundle:Voucher','v')
->innerJoin('v.shop', 's');
以下是生成的查询：
SELECT l0.* FROM voucher l0_ INNER JOIN shop_voucher l2_ ON l0_.id = l2_.voucher_id INNER JOIN shop l1_ ON l1_.id = l2_.shop_id;
所以我想知道为什么当我添加更多连接时系统没有找到正确的关系。
这是我与Many2Many关系的主要实体：
Shop.php
namespace AdminBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
use Doctrine\Common\Collections\ArrayCollection;
/**
* Shop.
*
* @ORM\Table(name="shop")
* @ORM\Entity(repositoryClass="AdminBundle\Entity\ShopRepository")
*/
class Shop
{
const SHOP_DEACTIVATED = 0;
const SHOP_ACTIVATED = 1;
const SHOP_HIGHLIGHTED_HOME = 1;
................................
/**
* @ORM\ManyToMany(targetEntity="Voucher", inversedBy="shop")
* @ORM\JoinTable(name="shop_voucher")
*/
private $voucher;
................................
Voucher.php
namespace AdminBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
#use Doctrine\Common\Collections\ArrayCollection;
/**
* Voucher.
*
* @ORM\Table(name="voucher")
* @ORM\Entity(repositoryClass="AdminBundle\Entity\VoucherRepository")
*/
class Voucher
{
................................
/**
* @ORM\ManyToMany(targetEntity="Shop", mappedBy="voucher", cascade={"persist"})
*/
private $shop;
................................
我已经检查过具有相同问题的堆栈，但我想知道如何扩展查询。我有我的问题的解决方案，如下所示，但没有得到上述场景中正在发生的事情。
$shopDataQuery = $connection->prepare('SELECT v.* FROM voucher AS v INNER JOIN shop_voucher AS sv ON sv.voucher_id = v.id INNER JOIN shop AS s ON s.id = sv.shop_id LEFT JOIN voucher_programs AS vp ON vp.id = v.program_id LEFT JOIN shop_history AS sh ON sh.shop = s.id WHERE s.shopStatus = :shopStatus AND s.highlightedHome = :highlightedHome AND s.offers = 'voucher' GROUP BY sh.shop ORDER BY v.discount_amount DESC LIMIT 6');
更新
这是shopHistory.php
namespace AdminBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
use Doctrine\Common\Collections\ArrayCollection;
/**
* shopHistory.
*
* @ORM\Table(name="shop_history")
* @ORM\Entity(repositoryClass="AdminBundle\Entity\shopHistoryRepository")
*/
class shopHistory
{
/**
* @var int
*
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @var int
* @ORM\ManyToOne(targetEntity="Shop", inversedBy="shopHistory")
* @ORM\JoinColumn(name="shop", referencedColumnName="id")
*/
private $shop;
这是VoucherProgram.php
namespace AdminBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
use Doctrine\Common\Collections\ArrayCollection;
/**
* VoucherProgram.
*
* @ORM\Table(name="voucher_program")
* @ORM\Entity(repositoryClass="AdminBundle\Entity\VoucherProgramRepository")
*/
class VoucherProgram
{
/**
* @var int
*
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @ORM\OneToMany(targetEntity="Voucher", mappedBy="program")
*/
private $voucher;
/**
* @ORM\OneToMany(targetEntity="Shop", mappedBy="vprogram")
*/
private $shop;
答案 0 :(得分：2)
我认为这是因为您已将您的关系命名为与您的字段相同：
SELECT v
FROM AdminBundle:Voucher v
INNER JOIN v.shop s
LEFT JOIN AdminBundle:VoucherPrograms vp ON vp.id = v.program_id
LEFT JOIN AdminBundle:shopHistory sh ON sh.shop = s.id
WHERE s.shopStatus = :shopStatus
AND s.highlightedHome = :highlightedHome
AND s.offers = 'voucher'
GROUP BY sh.shop
ORDER BY v.discount_amount DESC
尝试在 shopHistory 中重命名 shop_id 字段购物。
但我没有完整的实体模型来进行测试！
如果不是这样，请尝试粘贴5个实体的代码，请...
修改：
我使用symfony 2.8版本在本地计算机上进行测试。
我已经生成了给定模型的实体。
我刚刚在凭证实体中添加了这个：
/**
* @ORM\ManyToOne(targetEntity="voucherProgram", inversedBy="voucher")
*/
private $program;
我有这个疑问：
我的第一个查询
SELECT v
FROM AppBundle:Voucher v
INNER JOIN v.shop s
LEFT JOIN AppBundle:voucherProgram vp WITH vp.id = v.program
LEFT JOIN AppBundle:shopHistory sh
WITH sh.shop = s.id
GROUP BY sh.shop
我的第二个SQL查询
SELECT v0_.id AS id0, v0_.program_id AS program_id1
FROM voucher v0_
INNER JOIN shop_voucher s2_ ON v0_.id = s2_.voucher_id
INNER JOIN shop s1_ ON s1_.id = s2_.shop_id
LEFT JOIN voucher_program v3_ ON (v3_.id = v0_.program_id)
LEFT JOIN shop_history s4_ ON (s4_.shop = s1_.id)
GROUP BY s4_.shop
LIMIT 6
OFFSET 0
我认为生成的内容是正确的！
编辑2：
我尝试使用symfony标准版：
$shopData = $this->getDoctrine()
->getManager()
->createQueryBuilder()
->select('v')
->from('AppBundle:Voucher','v')
->innerJoin('v.shop', 's')
->leftJoin('AppBundle:voucherProgram', 'vp', 'WITH', 'vp.id = v.program')
->leftJoin('AppBundle:shopHistory', 'sh', 'WITH', 'sh.shop = s.id')
//->where('s.shopStatus = :shopStatus')
//->setParameter('shopStatus', Shop::SHOP_ACTIVATED)
//->andWhere('s.highlightedHome = :highlightedHome')
//->setParameter('highlightedHome', Shop::SHOP_HIGHLIGHTED_HOME)
//->andWhere('s.offers = \'voucher\'')
->setFirstResult(0)
->setMaxResults(6)
//->addOrderBy('v.discount_amount', 'DESC')
->groupBy('sh.shop')
->getQuery()
->getDql();
我的composer.json是：
"php": ">=5.3.9",
"symfony/symfony": "2.8.*",
"doctrine/orm": "^2.4.8",
"doctrine/doctrine-bundle": "~1.4",
"symfony/swiftmailer-bundle": "~2.3",
"symfony/monolog-bundle": "~2.4",
"sensio/distribution-bundle": "~5.0",
"sensio/framework-extra-bundle": "^3.0.2",
"incenteev/composer-parameter-handler": "~2.0"
我的实体：
<强>商店
namespace AppBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* Shop
*
* @ORM\Table(name="shop")
* @ORM\Entity(repositoryClass="AppBundle\Repository\ShopRepository")
*/
class Shop
{
const SHOP_DEACTIVATED = 0;
const SHOP_ACTIVATED = 1;
const SHOP_HIGHLIGHTED_HOME = 1;
/**
* @var int
*
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* Get id
*
* @return integer
*/
public function getId()
{
return $this->id;
}
/**
* @ORM\ManyToMany(targetEntity="Voucher", inversedBy="shop")
* @ORM\JoinTable(name="shop_voucher")
*/
private $voucher;
}
<强> shopHistory
namespace AppBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* shopHistory
*
* @ORM\Table(name="shop_history")
* @ORM\Entity(repositoryClass="AppBundle\Repository\shopHistoryRepository")
*/
class shopHistory
{
/**
* @var int
*
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* Get id
*
* @return integer
*/
public function getId()
{
return $this->id;
}
/**
* @var int
* @ORM\ManyToOne(targetEntity="Shop", inversedBy="shopHistory")
* @ORM\JoinColumn(name="shop", referencedColumnName="id")
*/
private $shop;
}
<强>券
namespace AppBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* Voucher
*
* @ORM\Table(name="voucher")
* @ORM\Entity(repositoryClass="AppBundle\Repository\VoucherRepository")
*/
class Voucher
{
/**
* @var int
*
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* Get id
*
* @return integer
*/
public function getId()
{
return $this->id;
}
/**
* @ORM\ManyToMany(targetEntity="Shop", mappedBy="voucher", cascade={"persist"})
*/
private $shop;
/**
* @ORM\ManyToOne(targetEntity="voucherProgram", inversedBy="voucher")
*/
private $program;
}
<强> voucherProgram
namespace AppBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* voucherProgram
*
* @ORM\Table(name="voucher_program")
* @ORM\Entity(repositoryClass="AppBundle\Repository\voucherProgramRepository")
*/
class voucherProgram
{
/**
* @var int
*
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @ORM\OneToMany(targetEntity="Voucher", mappedBy="program")
*/
private $voucher;
/**
* @ORM\OneToMany(targetEntity="Shop", mappedBy="vprogram")
*/
private $shop;
/**
* Get id
*
* @return integer
*/
public function getId()
{
return $this->id;
}
}
答案 1 :(得分：1)
据我所知，第一个给定的查询是DQL查询，第二个是SQL查询。 DQL查询可以更短，并且仍然可以按预期工作，因为它不运行表而是有效。因此，您需要较少的连接来加入m2m关系（在生成final SQL时会扩展为双连接）
如果使用给定的查询构建器生成DQL，我发现生成的DQL没有任何问题。