我有以下选择一个查询生成器:Symfony的学说的QueryBuilder添加where子句上加入
$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“首发”,但里面Starter.php我有一个协会“引用”像这样:
/**
* @var ArrayCollection
*
* @ORM\ManyToMany(targetEntity="StarterReference", inversedBy="starters")
* @ORM\JoinTable(name="starters_references")
*/
protected $references;
因此,在我的查询结果中,我有“starters”和“starters_references”表。 1启动器有许多启动器参考。现在问题是我需要选择的不是所有的启动器引用,而只是那些在名为“ref_usage”的列中有一些值的启动器引用。
所以我需要写在where子句中我的查询,我想这样的:
->where('reference.ref_usage = 1')
不过这样一来,我得到只有一个“启动”项与所有已列入参考文献。我需要所有初级项目,但只有参考ref_usage 1.
任何想法?
这是我使用的完整文件和功能。
控制器功能: http://pastebin.com/0tTEcQbn
实体 “Starter.php”: http://pastebin.com/BFLpKtec
实体 “StarterReference.php”: http://pastebin.com/Kr9pEMEW
编辑: 这是我的查询如果我使用:
->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
正如你可以看到它增加了ref_usage = 1 where子句。但问题是我不需要它,我只需要在内部加入我的引用时检查ref_usage。
完整查询构建器请 – Hooli
添加了我正在使用的整个文件。发布编辑。 – The50