Question

我有一个查询构建器，其中包含以下选择：

$starterRepository = $this->getDoctrine()
                          ->getManager()
                          ->getRepository('CatalogBundle:Starter');

$query = $starterRepository->createQueryBuilder('s')
                           ->where('s.active = 1')
                           ->orderBy('s.voltage, s.power, s.serial');

它选择表＆＃34; starters＆＃34;，但在Starter.php中我有一个关联＆＃34;引用＆＃34;像这样：

 /**
 * @var ArrayCollection
 *
 * @ORM\ManyToMany(targetEntity="StarterReference", inversedBy="starters")
 * @ORM\JoinTable(name="starters_references")
 */
protected $references;

所以在我的查询结果中，我有两个＆＃34; starters＆＃34;和＆＃34; starters_references＆＃34;表。 1个起动器有许多起动器参考。现在的问题是我需要选择不是所有的起始引用，而只选择名为＆＃34; ref_usage＆＃34;的列中具有某些值的引用。

所以我需要在查询中编写where子句，我正在尝试这个：

->where('reference.ref_usage = 1')

但是这样我只得到一个＆＃34;起动器＆＃34;包含所有引用的项目。我需要所有的初学者项目，但只有ref_usage 1的引用。

有什么想法吗？

以下是我正在使用的完整文件和功能。

控制器功能： http://pastebin.com/0tTEcQbn

实体＆＃34; Starter.php＆＃34;： http://pastebin.com/BFLpKtec

实体＆＃34; StarterReference.php＆＃34;： http://pastebin.com/Kr9pEMEW

编辑：这是我使用的查询：

->where('reference.ref_usage = 1')

SELECT COUNT(*) AS dctrn_count FROM (SELECT DISTINCT id0 FROM (SELECT s0_.id AS id0, s0_.serial AS serial1, s0_.voltage AS voltage2, s0_.power AS power3, s0_.rotation AS rotation4, s0_.teeth AS teeth5, s0_.module AS module6, s0_.b_terminal AS b_terminal7, s0_.comment AS comment8, s0_.commenten AS commenten9, s0_.commentru AS commentru10, s0_.commentpl AS commentpl11, s0_.commentde AS commentde12, s0_.commentes AS commentes13, s0_.type AS type14, s0_.adaptation AS adaptation15, s0_.alternative_product_1 AS alternative_product_116, s0_.alternative_product_2 AS alternative_product_217, s0_.alternative_product_3 AS alternative_product_318, s0_.alternative_product_4 AS alternative_product_419, s0_.active AS active20 FROM starters s0_ INNER JOIN starters_references s2_ ON s0_.id = s2_.starter_id INNER JOIN starter_reference s1_ ON s1_.id = s2_.starterreference_id WHERE s0_.active = 1 AND s1_.ref_usage = 1 ORDER BY s0_.voltage ASC, s0_.power ASC, s0_.serial ASC) dctrn_result) dctrn_table

如您所见，它将ref_usage = 1添加到where子句。但问题是我在这里不需要它，我只需要在内部加入我的引用时检查ref_usage。

Answer 1

您应该在查询中加入引用，然后添加where子句。

$query = $starterRepository->createQueryBuilder('s')
                       ->join('s.references', 'reference')
                       ->where('s.active = 1')
                       ->andwhere('reference.ref_usage = 1')
                       ->orderBy('s.voltage, s.power, s.serial');

Symfony Doctrine QueryBuilder在Join

1 个答案: