<VirtualHost *:80>
app.form.type.emergency:
class: AppBundle\Form\Type\ApproverType
arguments:
- "@service_container"
tags:
- { name: form.type }Symfony\Component\DependencyInjection\ContainerInterface/**
* @param ContainerInterface $container
*/
public function __construct(ContainerInterface $container)
{
$this->user = $container->get('security.token_storage')->getToken()->getUser();
}$search_articles = mysqli_prepare($connect, "SELECT post_id, post_title, post_description, post_keywords, post_author, post_date, post_image, post_content, post_link FROM post WHERE post_title LIKE ?");
mysqli_stmt_bind_param($search_articles, 's', "%$search%");require_once('database.php');
use Illuminate\Database\Capsule\Manager as Capsule;
class Checkers {
protected static $tables = [
'dnsrec' => 'dns',
.... etc.
];
public function check ($userid,$id , $type,$what) {
$query = Capsule::table(self::$tables[$type]) ->where('unid' , $id);
if ($what==='belong') {
$result = $query->where('userid', $userid)->first();
} else if ($what==='exist') {
$result = $query->first();
}
if (is_empty($result)) {
return true;
}
return false;
}
}foreach ($routes as $key => $value) {
if (Yii::$app->user->can($key)) {
return $this->redirect([$value]);
}
}Возможно ли как-то для поля задать псевдоним через as?
