$query = "SELECT a.*, cc.name AS category, dd.ezcity AS proploc, ee.name AS statename, ff.name AS cnname, ss.dealer_name AS propseller, u.name AS editor" . "\n FROM #__ezrealty AS a" . "\n LEFT JOIN #__ezrealty_catg AS cc ON cc.id = a.cid" . "\n LEFT JOIN #__ezrealty_locality AS dd ON dd.id = a.locid" . "\n LEFT JOIN #__ezrealty_state […]
Я делаю этот простой сайт, и я столкнулся с этой ошибкой: Моя функция: <?php function user_exists($username) { $username = sanitize($username); $query = mysqli_query($connect, "SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username'"); return (mysqli_result($query, === 0) 1) ? true : false; } ?> Мой журнал ошибок php: PHP Parse error: syntax error, unexpected '===' (T_IS_IDENTICAL) in […]