Database query with like condition

Question

Currently I am using the following query to select the data from database

$result = db_query("select data from {config} where name like '%IPPOOL_POOL%' ");
while ($row = $result->fetchObject()) {
  $ips=$row->value;
}

But this generated an error like

Undefined property: stdClass::$value

Please help me to implement same in Drupal 8.

Answer 1

According to Drupal db_like doc page

Get a database connection injected into your service from the container and call escapeLike() on it.

For example,

$injected_database->escapeLike($string);

Try something like:

$database = \Drupal::database();
$result = $database->select('table_name', 't')
    ->fields('t', ['field_1', 'field_2'])
    ->condition('field_1', "%" . $database->escapeLike($string) . "%", 'LIKE')
    ->execute()
    ->fetchAll();

Read Database api in Drupal 8 for a better understanding of database query in drupal 8.

Answer 2

If you are interested in syntax for query() (not for condition()) then here how to use like:

$db = \Drupal::database();
$oldTablesPrefix = $db->escapeLike('old_') . '%';
$oldTables = $db->query('SHOW TABLES LIKE :prefix', [':prefix' => $oldTablesPrefix])
  ->fetchCol();