doujiena0025
2017-06-02 09:19
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在symfony / doctrine中查询not null OneToMany关系

I am using Symfony 3.2. I have an entity called Site which contains a OneToMany relationship to another entity called Indication. 

class Site
{
    /**
     * @ORM\OneToMany(targetEntity="IhkBundle\Entity\Indication", mappedBy="site")
     */
    private $indications;
}

class Indication
{
    /**
     * @ORM\ManyToOne(targetEntity="IhkBundle\Entity\Site", inversedBy="indications")
     * @ORM\JoinColumn(name="site_id", referencedColumnName="id")
     */
    private $site;
}

I want to query for all sites where Indications are available. I only get an ArrayCollection which I don't know what to do with.

$sites = $repository->findAll();
foreach ($sites as $site) {
   $site->getIndications();
}

I also tried to use the queryBuilder like in this answer.

$query = $repository->createQueryBuilder('s');
$result = $query->where('s.indications IS NOT NULL')
    ->getQuery()
    ->getResult();

which throws the following error: 

[Semantical Error] line 0, col 46 near 'indications IS': Error: Invalid PathExpression. StateFieldPathExpression or SingleValuedAssociationField expected.

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我正在使用Symfony 3.2。 我有一个名为 Site 的实体，它包含一个名为 Indication 的另一个实体的OneToMany关系。 

 类Site 
 {
 / ** 
 * @ORM \ OneToMany（targetEntity =”IhkBundle \ Entity \ Indication“，mappedBy =”site“）
 * / 
 private $ indication  ; 
} 
 
class指示
 {
 / ** 
 * @ORM \ ManyToOne（targetEntity =“IhkBundle \ Entity \ Site”，inversedBy =“indication”）
 * @ORM \ JoinColumn（name  =“site_id”，referencedColumnName =“id”）
 * / 
 private $ site; 
} 
   
 
 
我想查询所有 站点，其中 Indications 可用。 我只得到一个我不知道该怎么做的ArrayCollection。 
 
 
  $ sites = $ repository-＆gt; findAll（  ）; 
foreach（$ sites as $ site）{
 $ site-＆gt; getIndications（）; 
} 
   
 
 
我也尝试使用queryBuilder 比如这个答案。 
 
 
   $ query = $ repository-＆gt; createQueryBuilder（'s'）; 
 $ result = $ query-＆gt; where（'s.indications IS NOT NULL'）
  - ＆gt; getQuery（）
  -  ＆gt; getResult（）; 
   
 
 
会引发以下错误： 
 
 
   [Semantical Error]第0行，col 46附近'指示IS'：错误：无效的PathExpression。 期望StateFieldPathExpression或SingleValuedAssociationField。
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1条回答 默认 最新

  • dongqiongjiong4740 2017-06-02 09:28
    已采纳

    Seems like this can be solved by SQL by doing a simple join, something like this should work:

    $query = $repository->createQueryBuilder('s');
$result = $query->select('p')
    ->join('p.indications', 'i')
    ->getQuery()
    ->getResult();
    已采纳该答案
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