При работе с базой данных выдаёт пустой результат, точнее выдаёт ошибку:
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, null given in Z:\home\admin.uz\www\second.php on line 20
Warning: mysqli_error() expects exactly 1 parameter, 0 given in Z:\home\admin.uz\www\second.php on line 20
Вот код:
//1.Create a db connection
define("DB_SERVER", "localhost");
define("DB_USER", "pavel");
define("DB_PASS", "2604545");
define("DB_NAME", "inhaur");
$connection = mysqli_connect(DB_SERVER,DB_USER,DB_PASS,DB_NAME);
//Test if connection succed
if(mysqli_connect_errno())
{
die("Database connection failed: ".
mysqli_connect_error().
"(" . mysqli_connect_errno(). ")"
);
}
$query = "SELECT 'id' FROM 'pages'";
$result = mysqli_query($connection,$query);
$arrayOf = mysqli_fetch_assoc($result) or die("ERROR: ".mysqli_error());