doctrine queryBuilder where IN collection - symfony

Question

On my entity I have an array collection of users
/**
* #ORM\ManyToMany(targetEntity="\UserBundle\Entity\User", mappedBy="acr_groups")
*/
protected $users;
public function __construct() {
$this->users = new \Doctrine\Common\Collections\ArrayCollection();
}
In my FormType I want to filter out those groups wherein current user is a member:
$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'
))
My problem is obviously on this line:
$qb->where(':user IN (g.users)');
How can I use my collection of users as the argument for the IN()?

Answer 1

Try below code
$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'
))
I have tried it in symfony 2.3 with doctrine2. You can use select function with createQueryBuilder() to get specific columns.

Answer 2

$q = $this->createQueryBuilder('v')
->select('v')
->andWhere('v.workingHours IN (:workingHours)')
->setParameter('workingHours', $workingHours);
From : Doctrine 2 WHERE IN clause using a collection of entities
Or according to doctrine documentation : http://docs.doctrine-project.org/projects/doctrine-orm/en/latest/reference/query-builder.html#the-expr-class
To insert IN condition in query builder with doctrine you can use 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'));
Syntaxe of IN :
$qb->expr()->in('u.id', array(1, 2, 3))
Also, Make sure that you do NOT use something similar to $qb->expr()->in('value', array('stringvalue')) as this will cause Doctrine to throw an Exception. Instead, use $qb->expr()->in('value', array('?1')) and bind your parameter to ?1

Answer 3

I ended up turning thing around a bit after unsuccessfully attempting some of your solutions. I manually created an array of the IDs I wanted.
There is probably a native way of doing this, seems like a pretty standard thing... this works however.
// 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'
))