[SOLVED] doctrine2 dql, use setParameter with % wildcard when doing a like comparison

doctrine2 dql, use setParameter with % wildcard when doing a like comparison

I want to use the parameter place holder - e.g. ?1 - with the % wild cards. that is, something like: "u.name LIKE %?1%" (though this throws an error). The docs have the following two examples: 1.

// Example - $qb->expr()->like('u.firstname', $qb->expr()->literal('Gui%'))
public function like($x, $y); // Returns Expr\Comparison instance

I do not like this as there is no protection against code injection.

// $qb instanceof QueryBuilder

// example8: QueryBuilder port of: "SELECT u FROM User u WHERE u.id = ?1 OR u.nickname LIKE ?2 ORDER BY u.surname DESC" using QueryBuilder helper methods
$qb->select(array('u')) // string 'u' is converted to array internally
   ->from('User', 'u')
   ->where($qb->expr()->orx(
       $qb->expr()->eq('u.id', '?1'),
       $qb->expr()->like('u.nickname', '?2')
   ))
   ->orderBy('u.surname', 'ASC'));

I do not like this because I need to search for terms within the object's properties - that is, I need the wild cards on either side.

Solution

When binding parameters to queries, DQL pretty much works exactly like PDO (which is what Doctrine2 uses under the hood).

So when using the LIKE statement, PDO treats both the keyword and the % wildcards as a single token. You cannot add the wildcards next to the placeholder. You must append them to the string when you bind the params.

$qb->expr()->like('u.nickname', '?2')
$qb->getQuery()->setParameter(2, '%' . $value . '%');

See this comment in the PHP manual.