Symfony Doctrine 查询生成器 LIKE 在 PostgreSQL 中不起作用

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.