doctrine queryBuilder where IN 集合

Question

在我的实体上，我有一个用户数组集合

/**
 * @ORM\ManyToMany(targetEntity="\UserBundle\Entity\User", mappedBy="acr_groups")
 */
protected $users;

public function __construct() {
    $this->users = new \Doctrine\Common\Collections\ArrayCollection();
}

在我的 FormType 中，我想过滤掉当前用户所属的那些组：

    $builder
    ->add('acr_group', EntityType::class, array(
        'label' => 'ATS',
        'class' => 'HazardlogBundle:ACRGroup',
        'query_builder' => function (EntityRepository $er) use ($user) { // 3. use the user variable in the querybilder
                $qb = $er->createQueryBuilder('g');
                $qb->where(':user IN (g.users)');
                $qb->setParameters( array('user' => $user) );
                $qb->orderBy('g.name', 'ASC');
                return $qb;
        },
        'choice_label' => 'name'
    ))

我的问题显然在这一行：

$qb->where(':user IN (g.users)');

如何使用我的用户集合作为 IN() 的参数？

Answer 1

$q = $this->createQueryBuilder('v')
    ->select('v')
    ->andWhere('v.workingHours IN (:workingHours)')
    ->setParameter('workingHours', $workingHours);

发件人：Doctrine 2 WHERE IN clause using a collection of entities

或者根据学说文档：http://docs.doctrine-project.org/projects/doctrine-orm/en/latest/reference/query-builder.html#the-expr-class 要使用 doctrine 在查询构建器中插入 IN 条件，您可以使用 expr()

$qb->add('select', new Expr\Select(array('u')))
   ->add('from', new Expr\From('User', 'u'))
   ->add('where', $qb->expr()->orX(
       $qb->expr()->eq('u.id', '?1'),
       $qb->expr()->like('u.nickname', '?2')
   ))
   ->add('orderBy', new Expr\OrderBy('u.name', 'ASC'));

IN 的语法：

$qb->expr()->in('u.id', array(1, 2, 3))

此外，请确保您不使用类似于 $qb->expr()->in('value', array('stringvalue')) 的东西，因为这会导致 Doctrine 抛出异常。相反，使用 $qb->expr()->in('value', array('?1')) 并将您的参数绑定到 ?1

Answer 2

试试下面的代码

$user = array(12,211,612,84,63,23); // Assuming User Ids whose groups you want to retrive

$builder
->add('acr_group', EntityType::class, array(
    'label' => 'ATS',
    'class' => 'HazardlogBundle:ACRGroup',
    'query_builder' => function (EntityRepository $er) use ($user) { 
            $qb = $er->createQueryBuilder('g');
            $qb->innerJoin('g.users', 'u'); // Inner Join with users
            $qb->where('u.id IN (:user)');
            $qb->setParameters( array('user' => $user) );
            $qb->orderBy('g.name', 'ASC');
            return $qb;
    },
    'choice_label' => 'name'
))

我已经在 symfony 2.3 和 doctrine2 中尝试过了。您可以使用 select 函数和 createQueryBuilder() 来获取特定的列。

Answer 3

在尝试您的一些解决方案失败后，我最终扭转了局面。我手动创建了一个我想要的 ID 数组。

可能有一种本机的方法可以做到这一点，这似乎是一种非常标准的方法……但是这很有效。

// 1. to inject user entity into this builder first make a construct function (remember to inject it from controller!)

function __construct($user)
{
    $this->user = $user;
}

/**
 * {@inheritdoc}
 */


public function buildForm(FormBuilderInterface $builder, array $options)
{
    $user = $this->user; // 2. instantiate the variable we created in our construct above

    //create group list array
    $groupList = $this->user->getACRGroups();
    $gla = array(); 
    foreach ($groupList as $g) {
        $gla[] = $g->getId();
    };

    $builder
    ->add('acr_group', EntityType::class, array(
        'label' => 'ATS',
        'class' => 'HazardlogBundle:ACRGroup',
        'query_builder' => function (EntityRepository $er) use ($gla) { // 3. use the user variable in the querybilder
                $qb = $er->createQueryBuilder('g');
                $qb->where('g.id IN (:gla)');
                $qb->setParameters( array('gla' => $gla) );
                $qb->orderBy('g.name', 'ASC');
                return $qb;
        },
        'choice_label' => 'name'
    ))

doctrine queryBuilder where IN 集合

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