php - 主义Mysql : multiple joins in query builder

Question

我有实体:Tag、User 和 Service。 Tag 与User 和Service 都是多对多关系。

 class Service{
      /**
       * @ORM\ManyToMany(targetEntity="Tag", mappedBy="serviceList")
       */
       private $tagList;
 }

 class User{
       /**
        * @ORM\ManyToMany(targetEntity="Tag", mappedBy="userList")
        */
       private $tagList;
 }

 class Tag{
       /**
        * @ORM\ManyToMany(targetEntity="User", inversedBy="tagList")
        * @ORM\JoinTable(name="tags_users")
        */
       private $userList;

       /**
        * @ORM\ManyToMany(targetEntity="Service", inversedBy="tagList")
        * @ORM\JoinTable(name="tags_services")
        */
       private $serviceList;
 }

使用查询构建器，我怎样才能返回所有至少有 1 个公共(public)标签的给定 User 服务，即与用户相关的服务

这是我写的一个查询，但它不起作用

  $query = $em->getRepository('AppBundle:Service')
        ->createQueryBuilder('service')
        ->join('service.tagList', 'serviceTag')
        ->join('AppBundle:BasicUser', 'user')
        ->join('user.tagList', 'userTag')
        ->where('user.id = :id')
        ->andWhere('serviceTag.id = userTag.id')
        ->setParameter('id', $user->getId())
    ;

错误:

 [Syntax Error] line 0, col 104: Error: Expected Literal, got "JOIN" (500 Internal Server Error)

Answer 1

您的查询是错误的。我认为最简单的解决方案是:

 $query = $em->getRepository('AppBundle:Service')
             ->createQueryBuilder('service')
             ->innerJoin('service.tagList', 'tag')
             ->andWhere(':user MEMBER OF tag.userList')
             ->setParameter('user',$user)
             ;