Symfony Doctrine query builder LIKE not working in PostgreSQL - php

Question

So I have a UserRepository that contains the following code
namespace AppBundle\Repository;
use \Doctrine\ORM\EntityRepository;
class UserRepository extends EntityRepository
{
public function findByRole($role)
{
$qb = $this->_em->createQueryBuilder();
$qb->select('u')
->from($this->_entityName, 'u')
->where('u.roles LIKE :roles')
->setParameter('roles', '%"'.$role.'"%');
return $qb->getQuery()->getResult();
}
}
This seems to be working perfectly fine if my database is MySQL but if i change the database to PostgreSQL this query throws the following error
An exception occurred while executing 'SELECT p0_.id AS id_0,
p0_.username AS username_1, p0_.password AS password_2, p0_.is_active
AS is_active_3, p0_.roles AS roles_4, p0_.name AS name_5, p0_.street
AS street_6, p0_.city AS city_7, p0_.state AS state_8, p0_.zip_code AS
zip_code_9, p0_.phone_number AS phone_number_10, p0_.dob AS dob_11,
p0_.company_name AS company_name_12, p0_.company_slug AS
company_slug_13, p0_.company_logo AS company_logo_14,
p0_.company_details AS company_details_15, p0_.stripe_customer_id AS
stripe_customer_id_16, p0_.created_at AS created_at_17, p0_.updated_at
AS updated_at_18 FROM px_user p0_ WHERE p0_.roles LIKE ?' with params
["%\"ROLE_EMPLOYER\"%"]:
SQLSTATE[42883]: Undefined function: 7 ERROR: operator does not exist:
json ~~ unknown LINE 1: ...at AS updated_at_18 FROM px_user p0_ WHERE
p0_.roles LIKE $1 ^ HINT: No operator matches the given name and
argument type(s). You might need to add explicit type casts.
This is the first time I am working with PostgreSQL so I am not getting what the problem is. After playing around with it for a while if I change the generated query to by adding the following piece
WHERE
p0_.roles::text LIKE '%ROLE_EMPLOYER%'
Everything works fine. Note the ::text.
So now how can i add that to the query builder so it works with PostgreSQL as well.

Answer 1

I solved the problem with the module boldtrn/jsonb-bundle but it created an error depending on the version of Postgres used.
I also solved the issue without the module by a native query like this :
public function findEmailsByRole($role)
{
return $this->_em->getConnection()->executeQuery(
"SELECT email FROM public.utilisateur WHERE roles::text LIKE :role",
['role'=>'%"' . $role . '"%']
)->fetchAll();
}

Answer 2

You can create your Function Like this
PS: Im Working on PostgreSQL Too
public function findByRole($role) {
$qb = $this->getEntityManager()->createQueryBuilder();
$qb->select('u')
->from($this->getEntityName(), 'u')
->where("u.roles LIKE '%$role%'")
;
return $qb->getQuery()->getResult();
}

Answer 3

I solved the problem by using JsonbBundle.
Following steps I took to fix it
$ composer require "boldtrn/jsonb-bundle
Updated the config.yml by adding the following in its respective place.
doctrine:
dbal:
types:
jsonb: Boldtrn\JsonbBundle\Types\JsonbArrayType
mapping_types:
jsonb: jsonb
orm:
dql:
string_functions:
JSONB_AG: Boldtrn\JsonbBundle\Query\JsonbAtGreater
JSONB_HGG: Boldtrn\JsonbBundle\Query\JsonbHashGreaterGreater
JSONB_EX: Boldtrn\JsonbBundle\Query\JsonbExistence
Changed the roles property to type jsonb
And inside the repository the following query worked
$query = $this->getEntityManager()->createQuery("SELECT u FROM AppBundle:User u WHERE JSONB_HGG(u.roles , '{}') LIKE '%EMPLOYER%' ");
$users = $query->getResult();
return $users;
The credit goes to Doctrine query postgres json (contains) json_array

Answer 4

public function findByRole($role)
{
$qb = $this->getEntityManager()->createQueryBuilder();
$qb->select('u')
->from($this->getEntityName(), 'u')
->where($qb->expr()->like('u.roles', ':roles')
->setParameter('roles', $qb->expr()->literal('%'.$role.'%'));
return $qb->getQuery()->getResult();
}
Replaced your custom string DQL to the QueryBuilder syntax.
I don't know if it might be related to the syntax you've got within your like statement: '%".$var."%', which might bug it. Hope this helps you solve it.
Doctrine Querybuilder documentation like :
// Example - $qb->expr()->like('u.firstname', $qb->expr()->literal('Gui%'))
public function like($x, $y); // Returns Expr\Comparison instance

Answer 5

Looks like from the PostgreSQL documentation on LIKE, you might need to do this:
$qb = $this->_em->createQueryBuilder();
$qb->select('u')
->from($this->_entityName, 'u')
->where('u.roles LIKE :roles')
->setParameter('roles', '\'%'.$role.'%\'');
return $qb->getQuery()->getResult();
Essentially having single quotes outside the percent signs by escaping them. I'm not sure if that will work, but can you try it?

Answer 6

Building up on #Lou Zito answer:
If you are dealing with Postgres please learn the power of json/b operations!
Casting a json/b array into text for like comparison is not a good approach. Cast it into jsonb instead (or even better: change your setup / migration(s) to use jsonb fields directly instead <3)
You can use ? operator, or the contains operator #> as an example.
See this full list of jsonb operators available
Please be aware, that ? will get interpreted as doctrine placeholder, you need to use ?? to escape it!
public function findByRole()
{
return $query = $this->getEntityManager()
->getConnection()
->executeQuery(<<<'SQL'
SELECT id FROM public.user
WHERE roles::jsonb ?? :role
ORDER BY last_name, first_name
SQL,
['role' => User::ROLE_XYZ]
)->fetchAllAssociative();
}
Hint: As you can see in the example above, I usually extract roles as public string constants in the User entity for easy access. I strongly recommend that as best practice as well.