How do you build up a LIKE condition in a db_select() query?

Question

How do you build up a LIKE condition in a db_select() query?

I know how to do this with a db_query() API call, but is there a special syntax / method for doing this with db_select()?

Answer 1

After digging through Drupal Documentation found a solution in db_like API documentation page and SelectQuery:condition handler doc page..

For example

$result = db_select('field_data_field_name', 'f')->fields('f', array(
    'entity_id',
    'field_name_value'
))
->condition('entity_type', 'user')
->condition('bundle', 'user')
->condition('deleted', 0)
->condition('field_name_value', '%' . db_like($last_item) . '%', 'LIKE')
->distinct()
->range(0, 10)
->execute();

Condition to use for like query is

->condition('field_name_value', '%' . db_like($last_item) . '%', 'LIKE')

Answer 2

You can also use Drupal\Core\Database\Database when creating "LIKE" query. This is Drupal 8 alternate syntax since db_select() is deprecated.

$database = Database::getConnection();
$query = $database->select('TABLE NAME', 'u')
    ->fields('u', array('column1','column2'));
$query->condition('column1', '%'.$database->escapeLike($search_phrase) . '%', 'LIKE');

Or add multiples with OR query.

$DB_OR = $query->orConditionGroup()
  // find match anywhere in field
  ->condition('column1', '%' . $database->escapeLike($search_phrase) . '%', 'LIKE')

  // find match starting at beginning
  ->condition('column2', $database->escapeLike($search_phrase) . '%', 'LIKE');

  // find match at end of field
  ->condition('column1', '%' . $database->escapeLike($search_phrase), 'LIKE')
$query->condition($DB_OR); // Add OR object as condition
$result = $query->execute();

Answer 3

For using "like" in db_select is as below and worked for me...

$search_value = "testvalue@xmail.com";
$query = db_select('TABLE NAME', 'u');
$query->fields('u', ['ROW1', 'ROW2']);
$query->condition('ROW1', '%' . db_like($search_value) . '%', 'LIKE');
$result = $query->execute();

Here exact value in ROW1 is "testvalue@xmail.com" and by using LIKE condition in db_select I got the output as "testvalue@xmail.com".

Answer 4

it works in Drupal8 Version 1

$exp = "1";
$connection = \Drupal::database();
$query = $connection ->select('node', 'n');
    $query->fields('n');
    $query->condition('nid', '%'.db_like($exp), 'LIKE');
    $records = $query->execute();
foreach ($records as $record) {
ksm($record);
}

Version 2

$connection = \Drupal::database();
$query = $connection->query("SELECT *FROM {node} WHERE nid like '%' :nid '%'", [
  ':nid' => "1",
]);
$records = $query->fetchAll();

foreach ($records as $record) {
ksm($record);
}

Answer 5

Since db_like() is deprecated and was removed in Drupal 9, the documentation suggests to use Connection::escapeLike() instead.

Here is a working example for Drupal 8 and 9:

$connection = Database::getConnection();
$query = $connection->select('your_table_name', 'table_alias')
  ->fields('table_alias', ['whatever_field', 'other_field'])
  ->condition('my_field_name', '%' . $connection->escapeLike($yourSearchString) . '%', 'LIKE');

$result = $query->execute()->fetchAll();